Otdfbt

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.

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

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...

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 $5$.

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$$(x+3)^2+(y-1)^2 = 25$$

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.