how to find the equation of a circle given points of the circle [closed]Finding the equation of a circle from given points on it and line on which the centre lies.Finding the equation of a circle given two points on the circleHow to find the equation of diameter of a circle that passes through the origin?Find the equation of a hyperbola, given a point on it and the length of the transverse axisFind the equation of the circle which touches the $X$ axis at tIn the given figure, $R$ is the center of the circle…How to Find the centre of a circle, using two points on the circumference.Finding equation of circle under the given geometric conditionsHow do I find the equation to a tangent of a circle given a gradientFind the equation of the circle passing through two points (4,4) and (-2.-2) with radius 4

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how to find the equation of a circle given points of the circle [closed]


Finding the equation of a circle from given points on it and line on which the centre lies.Finding the equation of a circle given two points on the circleHow to find the equation of diameter of a circle that passes through the origin?Find the equation of a hyperbola, given a point on it and the length of the transverse axisFind the equation of the circle which touches the $X$ axis at tIn the given figure, $R$ is the center of the circle…How to Find the centre of a circle, using two points on the circumference.Finding equation of circle under the given geometric conditionsHow do I find the equation to a tangent of a circle given a gradientFind the equation of the circle passing through two points (4,4) and (-2.-2) with radius 4













4












$begingroup$


Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$



I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.



Edit: this post was marked as missing details/context, however I copied the question in the exact same way in which the question was given to me. I left out nothing and no additional details were given.










share|cite|improve this question











$endgroup$



closed as off-topic by Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby Apr 28 at 2:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:33










  • $begingroup$
    Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:35
















4












$begingroup$


Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$



I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.



Edit: this post was marked as missing details/context, however I copied the question in the exact same way in which the question was given to me. I left out nothing and no additional details were given.










share|cite|improve this question











$endgroup$



closed as off-topic by Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby Apr 28 at 2:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:33










  • $begingroup$
    Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:35














4












4








4





$begingroup$


Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$



I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.



Edit: this post was marked as missing details/context, however I copied the question in the exact same way in which the question was given to me. I left out nothing and no additional details were given.










share|cite|improve this question











$endgroup$




Find the equation of the circle which passes through points $P(-6,5)$, $Q(2,1)$ and has its centre lying on the line $y=x+4$



I would appreciate any working given, as I do not understand how to find the equation of the circle from this information alone.



Edit: this post was marked as missing details/context, however I copied the question in the exact same way in which the question was given to me. I left out nothing and no additional details were given.







geometry analytic-geometry circles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 29 at 13:17







injustice fellow

















asked Apr 27 at 16:29









injustice fellowinjustice fellow

414




414




closed as off-topic by Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby Apr 28 at 2:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby Apr 28 at 2:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, A. Goodier, A. Pongrácz, Paul Frost, Thomas Shelby
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:33










  • $begingroup$
    Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:35

















  • $begingroup$
    Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:33










  • $begingroup$
    Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
    $endgroup$
    – Mauro ALLEGRANZA
    Apr 27 at 16:35
















$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
Apr 27 at 16:33




$begingroup$
Use the equation of a circle with radius $r$ (unknown) and center $(x_0, y_0)$ (unknown).
$endgroup$
– Mauro ALLEGRANZA
Apr 27 at 16:33












$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
Apr 27 at 16:35





$begingroup$
Then solve te system of equations obtained from the previous one (3 unknown) knowing that points P and Q satisfy the equation of the circle and that the coordinates of the center satisfy the equation of the line.
$endgroup$
– Mauro ALLEGRANZA
Apr 27 at 16:35











5 Answers
5






active

oldest

votes


















6












$begingroup$

Hint:



Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.



The radius would be the distance from the centre to any of the given points.



P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)



P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.





Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points






share|cite|improve this answer











$endgroup$








  • 3




    $begingroup$
    +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
    $endgroup$
    – TonyK
    Apr 27 at 17:04










  • $begingroup$
    @TonyK yeah thanks for pointing it out. I'll edit it.
    $endgroup$
    – Eagle
    Apr 27 at 17:05










  • $begingroup$
    @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
    $endgroup$
    – Maria Mazur
    Apr 27 at 17:11










  • $begingroup$
    @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
    $endgroup$
    – TonyK
    Apr 27 at 17:17


















3












$begingroup$

The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:



$$(x-a)^2+(y-a-4)^2 = r^2$$



Now put both points in to this equation and you get:



$$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$



and



$$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$



Now solve this system...






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Let the center be at $(t,t+4)$. We express that the circle is through the given points:



    $$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$



    After simplification,



    $$8t+24=0$$ and you are nearly done.




    $t=-3$, center $(-3,1)$, radius $5$.







    share|cite|improve this answer











    $endgroup$




















      1












      $begingroup$

      Let the center be $C(x_0,y_0)$



      As P and Q lie on the circle, CP = CQ = radius of the circle.
      $$CP = CQ$$
      $$CP^2 = CQ^2$$
      $$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
      $$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
      $$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
      $$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
      We also have$$y_1 = x_1+4$$
      So,$$2x_1 - x_1 - 4+ 7 = 0 $$
      $$x_1 = -3$$
      $$y_1 = x_1+4 = 1$$
      Centre $C = C(-3,1)$



      Radius $r = CQ = sqrt(-3-2)^2 + (1-1)^2 = 5$$



      Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
      or$$(x+3)^2+(y-1)^2 = 25$$






      share|cite|improve this answer









      $endgroup$




















        0












        $begingroup$

        Consider the circunference $C$
        $$
        (x + 3)^2 + (y - 1)^2 = 25
        $$

        We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.






        share|cite|improve this answer









        $endgroup$



















          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Hint:



          Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.



          The radius would be the distance from the centre to any of the given points.



          P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)



          P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.





          Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points






          share|cite|improve this answer











          $endgroup$








          • 3




            $begingroup$
            +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
            $endgroup$
            – TonyK
            Apr 27 at 17:04










          • $begingroup$
            @TonyK yeah thanks for pointing it out. I'll edit it.
            $endgroup$
            – Eagle
            Apr 27 at 17:05










          • $begingroup$
            @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
            $endgroup$
            – Maria Mazur
            Apr 27 at 17:11










          • $begingroup$
            @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
            $endgroup$
            – TonyK
            Apr 27 at 17:17















          6












          $begingroup$

          Hint:



          Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.



          The radius would be the distance from the centre to any of the given points.



          P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)



          P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.





          Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points






          share|cite|improve this answer











          $endgroup$








          • 3




            $begingroup$
            +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
            $endgroup$
            – TonyK
            Apr 27 at 17:04










          • $begingroup$
            @TonyK yeah thanks for pointing it out. I'll edit it.
            $endgroup$
            – Eagle
            Apr 27 at 17:05










          • $begingroup$
            @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
            $endgroup$
            – Maria Mazur
            Apr 27 at 17:11










          • $begingroup$
            @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
            $endgroup$
            – TonyK
            Apr 27 at 17:17













          6












          6








          6





          $begingroup$

          Hint:



          Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.



          The radius would be the distance from the centre to any of the given points.



          P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)



          P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.





          Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points






          share|cite|improve this answer











          $endgroup$



          Hint:



          Since the centre of the circle will also lie on the perpendicular bisector of the given two points, so you can find the coordinates as the intersection of the bisector and $y=x+4$.



          The radius would be the distance from the centre to any of the given points.



          P.S. In this case, the perpendicular bisector is different from the given line. If it weren't so, then the given points would be the ends of a diameter, from which you could easily find the center lying midway.(Thanks to @TonyK for pointing out the inaccuracy)



          P.S. If the perpendicular bisector is the given line, then there won't be a unique circle satisfying the conditions, but you'll get a family of circles.





          Image for the P.S. part, showing two of the possible circles if the given line was also a perpendicular bisector of the given points







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 27 at 17:24

























          answered Apr 27 at 16:45









          EagleEagle

          22713




          22713







          • 3




            $begingroup$
            +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
            $endgroup$
            – TonyK
            Apr 27 at 17:04










          • $begingroup$
            @TonyK yeah thanks for pointing it out. I'll edit it.
            $endgroup$
            – Eagle
            Apr 27 at 17:05










          • $begingroup$
            @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
            $endgroup$
            – Maria Mazur
            Apr 27 at 17:11










          • $begingroup$
            @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
            $endgroup$
            – TonyK
            Apr 27 at 17:17












          • 3




            $begingroup$
            +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
            $endgroup$
            – TonyK
            Apr 27 at 17:04










          • $begingroup$
            @TonyK yeah thanks for pointing it out. I'll edit it.
            $endgroup$
            – Eagle
            Apr 27 at 17:05










          • $begingroup$
            @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
            $endgroup$
            – Maria Mazur
            Apr 27 at 17:11










          • $begingroup$
            @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
            $endgroup$
            – TonyK
            Apr 27 at 17:17







          3




          3




          $begingroup$
          +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
          $endgroup$
          – TonyK
          Apr 27 at 17:04




          $begingroup$
          +1 for being the only answer so far to use the geometry of the thing. But your PS is wrong: if the perpendicular bisector is coincident with the given line, then the centre of the circle can be any point on the line.
          $endgroup$
          – TonyK
          Apr 27 at 17:04












          $begingroup$
          @TonyK yeah thanks for pointing it out. I'll edit it.
          $endgroup$
          – Eagle
          Apr 27 at 17:05




          $begingroup$
          @TonyK yeah thanks for pointing it out. I'll edit it.
          $endgroup$
          – Eagle
          Apr 27 at 17:05












          $begingroup$
          @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
          $endgroup$
          – Maria Mazur
          Apr 27 at 17:11




          $begingroup$
          @TonyK And how do we get the equation $$(x-a)^2+(y-b)^2 = r^2$$ without geometry?
          $endgroup$
          – Maria Mazur
          Apr 27 at 17:11












          $begingroup$
          @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
          $endgroup$
          – TonyK
          Apr 27 at 17:17




          $begingroup$
          @MariaMazur: What I really meant was that you can solve this problem simply by finding the intersection of two straight lines. There is no need for quadratic equations.
          $endgroup$
          – TonyK
          Apr 27 at 17:17











          3












          $begingroup$

          The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:



          $$(x-a)^2+(y-a-4)^2 = r^2$$



          Now put both points in to this equation and you get:



          $$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$



          and



          $$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$



          Now solve this system...






          share|cite|improve this answer









          $endgroup$

















            3












            $begingroup$

            The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:



            $$(x-a)^2+(y-a-4)^2 = r^2$$



            Now put both points in to this equation and you get:



            $$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$



            and



            $$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$



            Now solve this system...






            share|cite|improve this answer









            $endgroup$















              3












              3








              3





              $begingroup$

              The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:



              $$(x-a)^2+(y-a-4)^2 = r^2$$



              Now put both points in to this equation and you get:



              $$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$



              and



              $$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$



              Now solve this system...






              share|cite|improve this answer









              $endgroup$



              The equation of circle is $$(x-a)^2+(y-b)^2 = r^2$$where $(a,b)$ is a center of a circle and $r$ radius of a circle. Since ceneter is on a line $y=x+4$ we have $b=a+4$ so we have now:



              $$(x-a)^2+(y-a-4)^2 = r^2$$



              Now put both points in to this equation and you get:



              $$ P: ;;;(-6-a)^2+(1-a)^2 = r^2$$



              and



              $$ Q: ;;;(2-a)^2+(-a-3)^2 = r^2$$



              Now solve this system...







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Apr 27 at 16:53









              Maria MazurMaria Mazur

              51.1k1362129




              51.1k1362129





















                  2












                  $begingroup$

                  Let the center be at $(t,t+4)$. We express that the circle is through the given points:



                  $$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$



                  After simplification,



                  $$8t+24=0$$ and you are nearly done.




                  $t=-3$, center $(-3,1)$, radius $5$.







                  share|cite|improve this answer











                  $endgroup$

















                    2












                    $begingroup$

                    Let the center be at $(t,t+4)$. We express that the circle is through the given points:



                    $$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$



                    After simplification,



                    $$8t+24=0$$ and you are nearly done.




                    $t=-3$, center $(-3,1)$, radius $5$.







                    share|cite|improve this answer











                    $endgroup$















                      2












                      2








                      2





                      $begingroup$

                      Let the center be at $(t,t+4)$. We express that the circle is through the given points:



                      $$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$



                      After simplification,



                      $$8t+24=0$$ and you are nearly done.




                      $t=-3$, center $(-3,1)$, radius $5$.







                      share|cite|improve this answer











                      $endgroup$



                      Let the center be at $(t,t+4)$. We express that the circle is through the given points:



                      $$r^2=(t+6)^2+(t+4-5)^2=(t-2)^2+(t+4-1)^2.$$



                      After simplification,



                      $$8t+24=0$$ and you are nearly done.




                      $t=-3$, center $(-3,1)$, radius $5$.








                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Apr 27 at 18:37

























                      answered Apr 27 at 17:13









                      Yves DaoustYves Daoust

                      135k676234




                      135k676234





















                          1












                          $begingroup$

                          Let the center be $C(x_0,y_0)$



                          As P and Q lie on the circle, CP = CQ = radius of the circle.
                          $$CP = CQ$$
                          $$CP^2 = CQ^2$$
                          $$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
                          $$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
                          $$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
                          $$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
                          We also have$$y_1 = x_1+4$$
                          So,$$2x_1 - x_1 - 4+ 7 = 0 $$
                          $$x_1 = -3$$
                          $$y_1 = x_1+4 = 1$$
                          Centre $C = C(-3,1)$



                          Radius $r = CQ = sqrt(-3-2)^2 + (1-1)^2 = 5$$



                          Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
                          or$$(x+3)^2+(y-1)^2 = 25$$






                          share|cite|improve this answer









                          $endgroup$

















                            1












                            $begingroup$

                            Let the center be $C(x_0,y_0)$



                            As P and Q lie on the circle, CP = CQ = radius of the circle.
                            $$CP = CQ$$
                            $$CP^2 = CQ^2$$
                            $$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
                            $$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
                            $$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
                            $$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
                            We also have$$y_1 = x_1+4$$
                            So,$$2x_1 - x_1 - 4+ 7 = 0 $$
                            $$x_1 = -3$$
                            $$y_1 = x_1+4 = 1$$
                            Centre $C = C(-3,1)$



                            Radius $r = CQ = sqrt(-3-2)^2 + (1-1)^2 = 5$$



                            Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
                            or$$(x+3)^2+(y-1)^2 = 25$$






                            share|cite|improve this answer









                            $endgroup$















                              1












                              1








                              1





                              $begingroup$

                              Let the center be $C(x_0,y_0)$



                              As P and Q lie on the circle, CP = CQ = radius of the circle.
                              $$CP = CQ$$
                              $$CP^2 = CQ^2$$
                              $$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
                              $$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
                              $$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
                              $$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
                              We also have$$y_1 = x_1+4$$
                              So,$$2x_1 - x_1 - 4+ 7 = 0 $$
                              $$x_1 = -3$$
                              $$y_1 = x_1+4 = 1$$
                              Centre $C = C(-3,1)$



                              Radius $r = CQ = sqrt(-3-2)^2 + (1-1)^2 = 5$$



                              Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
                              or$$(x+3)^2+(y-1)^2 = 25$$






                              share|cite|improve this answer









                              $endgroup$



                              Let the center be $C(x_0,y_0)$



                              As P and Q lie on the circle, CP = CQ = radius of the circle.
                              $$CP = CQ$$
                              $$CP^2 = CQ^2$$
                              $$(x_1-(-6))^2 + (y_1 -5)^2 = (x_1-2)^2 + (y_1 -1)^2$$
                              $$x_1^2 + 12x_1 +36 + y_1^2 - 10y_1 +25 = x_1^2 -4x_1 +4 + y_1^2 - 2y_1 +1$$
                              $$12x_1+36 - 10y_1 +25 = -4x_1 + 4 -2y_1 + 1$$
                              $$16x_1 - 8y_1 +56 = 0$$ or $$ 2x_1 - y_1 + 7 =0$$
                              We also have$$y_1 = x_1+4$$
                              So,$$2x_1 - x_1 - 4+ 7 = 0 $$
                              $$x_1 = -3$$
                              $$y_1 = x_1+4 = 1$$
                              Centre $C = C(-3,1)$



                              Radius $r = CQ = sqrt(-3-2)^2 + (1-1)^2 = 5$$



                              Thus the equation is,$$(x-x_1)^2 + (y-y_1)^2 = r^2$$
                              or$$(x+3)^2+(y-1)^2 = 25$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Apr 27 at 16:57









                              Ak19Ak19

                              79810




                              79810





















                                  0












                                  $begingroup$

                                  Consider the circunference $C$
                                  $$
                                  (x + 3)^2 + (y - 1)^2 = 25
                                  $$

                                  We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.






                                  share|cite|improve this answer









                                  $endgroup$

















                                    0












                                    $begingroup$

                                    Consider the circunference $C$
                                    $$
                                    (x + 3)^2 + (y - 1)^2 = 25
                                    $$

                                    We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.






                                    share|cite|improve this answer









                                    $endgroup$















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Consider the circunference $C$
                                      $$
                                      (x + 3)^2 + (y - 1)^2 = 25
                                      $$

                                      We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Consider the circunference $C$
                                      $$
                                      (x + 3)^2 + (y - 1)^2 = 25
                                      $$

                                      We have that $P,Qin C$ (just replacing coordinates). And the center is $(-3,1)$ which is in the line $y=x+4$. Since $P,Qin C$ the center must be in the perpendicular line ($-2x+y=7$) to the one that joins $P,Q$. Since the intersection between the two lines that we have consider is a point, we deduce that the solution is unique and is the one we have proposed.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Apr 27 at 17:00









                                      davidivadfuldavidivadful

                                      14410




                                      14410













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