Find the line with a positive slope that is tangent to two circles

Question

There is a line with positive slope which is tangent to the circle x^2+y^2=1 at some point P and which is also tangent to the circle $(x-3)^2 + y^2 = 4$ at a point Q. Find the equation of this line.

Answer 1

There are some neat ways to think about this question. I strongly recommend you draw a picture.

Let $L_1$ be a line tangent to the first circle at a point $P_{12}$. Let $L_3$ be the line parallel to $L_1$ through the origin, and $L_2$ and $L_4$ be the lines perpendicular to $L_1$ through the origin and $(3,0)$ respectively. Let $P_{ij}$ be the intersection between lines $L_i$ and $L_j$. The four lines described form a rectangle, with side lengths 1 along $L_2$ and $L_4$. If $L_1$ is also tangent to the second circle, then the distance between $P_{34}$ and the center $(3,0)$ is equal to 1 (the radius 2 minus the side length 1 along $L_4$).

Now we actually compute with coordinates. Suppose the equation for $L_3$ is $y=mx$, so the equation for $L_4$ is $y = -\frac{1}{m}(x - 3)$. The intersection is $P_{34} = \left(\frac{3}{m^2+1}, \frac{3m}{m^2+1}\right)$. We can use the known distance between $P_{34}$ and $(3,0)$ to solve for $m$: \begin{align*} \sqrt{\frac{9m^4}{(m^2+1)^2} + \frac{9m^2}{(m^2+1)^2}} &= 1 \\ \frac{3m}{\sqrt{m^2+1}} &= 1 \\ 8m^2 &= 1 \\ m &= \frac{1}{2\sqrt{2}} \end{align*} Now $P_{12}$ is 1 unit along the line $y = -2\sqrt{2}x$ (in the 4th quadrant), so it's $\left(-\frac{1}{3}, \frac{2\sqrt{2}}{3}\right)$. Finally, $L_1$ has the equation $$y - \frac{2\sqrt{2}}{3} = \frac{1}{2\sqrt{2}}\left(x + \frac{1}{3}\right)$$

Answer 2

The tangent to the circle $x^2+y^2=1$ at some point $p(h,k)$ is $hx+ky=1$.

If this tangent is also a tangent to the circle ${(x-3)}^2+y^2=4$ then

The perpendicular distance of $hx+ky=1$ from the center of the circle ${(x-3)}^2+y^2=4$ = The radius of the circle ${(x-3)}^2+y^2=4$

i.e., $\frac{3h+0.k-1}{\sqrt{h^2+k^2}}= 2$

Squaring on both sides,

$\frac{{(3h-1)}^2}{h^2+k^2}=4$

But we know that $p:(h,k)$ lies on the circle $x^2+y^2=1$ therefore, $h^2+k^2=1$

The equation becomes, ${(3h-1)}^2=4$ or, $9h^2-6h+1=4$ or, $9h^2-6h-3=0$ $(h-1)(9h+3)=0$ i.e, $h=1,-\frac{1}{3}$ correspondingly $k= 0,\pm \frac{2\sqrt2 }{3}$

Therefore the point $p:(h,k)$ are $(1,0)$ or $(-\frac{1}{3},\frac{2\sqrt2 }{3} )$ or $(-\frac{1}{3},-\frac{2\sqrt2 }{3} )$

Now the slope of the tangent $hx+ky=1$ is $-\frac{h}{k}$ is given positive.

Therefore, $P:(h,k):=(-\frac{1}{3},\frac{2\sqrt2 }{3} )$

And the tangent of the equation is $-\frac{1}{3}x+\frac{2\sqrt2 }{3}y=1$ i.e., $x-2\sqrt2 y+3=0$