Otdfbt

Consider the two circles determined by $(x-1)^2 + y^2 = 1$ and $(x-2.5)^2 + y^2 = (1/2)^2$. Find the (explicit) equation of the line that lies tangent to both circles.

I have never seen a clean or clever solution to this problem. This problem came up once at a staff meeting for a tutoring center I worked at during undergrad. I recall my roommate and I - after a good amount of time symbol pushing - were able to visibly see a solution by inspection, then verify it by plugging in. I have never seen a solid derivation of a solution to this though, so I would like to see what MSE can come up with for this!

I took a short stab at it today before posting, and got that it would be determined by the solution to the equation $$left( fraccos(theta)sin(theta) + 2cos(theta) + 3 right)^2 -4left( fraccos(theta)^2sin(theta)^2+1 right)left(frac-cos(theta)^3sin(theta)^2+fraccos(theta)^4sin(theta)^2+3 right).$$

The solution $theta$ would then determine the line $$y(x) = frac-cos(theta)sin(theta)(x) + fraccos(theta)^2sin(theta) + sin(theta).$$

Not only do I not want to try and solve that, I don't even want to try expanding it out :/

As peterwhy points out in the comments, there are three tangent lines. By inspection, one is $x = 2$, as pointed out by J.W. Tanner in the comments.

The other two can be identified by similar triangles. Suppose that we have a line tangent to both circles, and let the points of tangency be $T_1$ and $T_2$. Let the circle centers be $O_1$ and $O_2$. Finally, let the point where this line intersects the $x$-axis be called $P$. Then $triangle PO_1T_1$ and $triangle PO_2T_2$ are similar (do you see why?). Since $O_1T_1 = 2O_2T_2$, we must have $PO_1 = 2PO_2$, and therefore $P$ must be at $(4, 0)$. Note that $PT_1 = sqrt3^2-1^2 = sqrt8$, and therefore our tangent line must have slope $pm frac1sqrt8$.

(For simplicity, I only show one of the tangent lines; the other is its mirror image across the $x$-axis.) From this, we get the equation of the two remaining tangent lines

$$
y = pm fracx-4sqrt8
$$

The equation of the second circle can be written as $x^2+y^2-5x+6=0$.

Let $P(h,k)$ be a point on the second circle.

Then the equation of the tangent to the second circle at $P$ is $hx+ky-dfrac52(x+h)+6=0$.

If it is also a tangent to the first circle, the distance from $(1,0)$ to this line is $1$.

beginalign*
fracsqrt(h-frac52)^2+k^2&=1\
fracsqrt(frac12)^2&=1\
-3h+7&=pm1
endalign*
So, we have $h=2$ or $h=frac83$.

If $h=2$, $k=pmsqrt(frac12)^2-(2-frac52)^2=0$ and the common tangent is $x-2=0$.

If $h=frac83$, $k=pmsqrt(frac12)^2-(frac83-frac52)^2=pmfracsqrt23$ and the common tangents are $xpm2sqrt2y-4=0$.

Use dual conics: If we represent a circle (in fact, any nondegenerate conic) with the homogeneous matrix $C$ so that its equation is $(x,y,1)C(x,y,1)^T=0$, lines $lambda x+mu y+tau=0$ tangent to the circle satisfy the dual conic equation $(lambda,mu,tau),C^-1(lambda,mu,tau)^T=0$. (This equation captures the fact that the pole of a tangent line lies on the line.)

The matrix that corresponds to the circle $(x-h)^2+(y-k)^2=r^2$ is $$C=beginbmatrix1&0&-h\0&1&-k\-h&-k&h^2+k^2-r^2endbmatrix$$ with inverse $$C^-1=frac1r^2beginbmatrixr^2-h^2&-hk&-h\-hk&r^2-k^2&-k\-h&-k&-1endbmatrix.$$ For the circles in this problem, the resulting dual equations are $$mu^2-tau^2-2lambdatau = 0 \ -24lambda^2+mu^2-4tau^2-20lambdatau = 0.$$ Both circles’ centers lie on the $x$-axis but their radii differ, so no common tangent is horizontal. Thus, we can set $lambda=1$ and solve the slightly simpler system to obtain the solutions $mu=0$, $tau=-2$ and $mu=pm2sqrt2$, $tau=-4$, i.e., the three common tangent lines are $$x=2 \ xpm2sqrt2 y=4.$$

This general method works for any pair of nondegenerate conics: it finds their common tangents by solving a dual problem of the intersection of two (possibly imaginary) conics. For a pair of circles, however, there’s a simple way to find common tangents via similar triangles, as demonstrated in Brian Tung’s answer.

To find a common tangent to two arbitrary circles, you can use the following trick: "deflate" the smaller circle (let $1$) so that it shrinks to a point, while the second shrinks to the radius $r_2-r_1$. Doing this, the direction of the tangent doesn't change and the tangency point forms a right triangle with the two centers.

By elementary trigonometry, the angle $phi$ between the axis through the centers and the tangent is drawn from

$$dsinphi=r_2-r_1,$$ where $d$ is the distance between the centers. The direction of the axis is such that

$$tantheta=fracy_2-y_1x_2-x_1.$$

So the equation of the tangent is given by

$$(x-x_1)sin(theta+phi)-(y-y_1)cos(theta+phi)=0.$$

The original tangent is a parallel at distance $r_1$, hence

$$(x-x_1)sin(theta+phi)-(y-y_1)cos(theta+phi)=r_1.$$

As already stated, there are three lines that can be drawn that are tangent to both circles. One is the trivial line $$x = 2,$$ which is tangent at the point $(2,0)$, which is also the common point of tangency of the two circles.

The other two lines are reflections of each other in the $x$-axis; these can be found by noting that a homothety that takes the point $(0,0)$ to $(2,0)$ and $(2,0)$ to $(3,0)$ will map the first circle to the second, and the center of this homothety will be a fixed point of this map.

To this end, let $(x',y') = (a x + b, a y + d)$, so we now solve the system $$(b, d) = (2,0) \ (2a + b, d) = (3,0).$$ That is to say, $b = 2$, $d = 0$, $a = 1/2$, and the desired homothety is $$(x',y') = (x/2 + 2, y/2).$$ The unique fixed point is found by setting $(x',y') = (x,y)$ from which we obtain $(x,y) = (4,0)$. Thus the tangent lines pass through this point and have the form $y = m(x - 4)$ where $m$ is the slope. Such a line will be tangent if the system of equations $$(x-1)^2 + y^2 = 1, \ y = m(x-4)$$ has exactly one solution. Eliminating $y$ gives the quadratic $$(m^2+1)x^2 - 2(4m^2+1)x + 16m^2 = 0,$$ for which the solution has a unique root if the discriminant is zero; i.e., $$0 = 4(4m^2+1)^2 - 4(m^2+1)(16m^2) = 4 - 32m^2.$$ Therefore, $m = pm 1/sqrt2$ and the desired lines are $$y = pm fracx-42 sqrt2,$$ in addition to the previous line $x = 2$ described.

Beside the obvious solution of $x=2$ you also have two more common tangent lines, namely $$y= pm frac sqrt 24 (x-4)$$

The two non-obvious tangent lines pass through $(4,0)$ and their y-intercepts are respectively $(0,pm sqrt 2)$