Coordinates of a meetpoint of two tangent (fun)

Question

A line is a tangent to the circle $C_1:x^2+y^2+4x-16y+23=0$, at the point$(-5,14)$. This line cuts another Circle $C_2:x^2+y^2+56x+10y+744=0$, at points $A$ and $B$. The tangents to $C_2$ at $A$ and $B$ meet at point $C$. Find the coordinates of $C$.

Answer 1

I think that explicit computations is the way in this question.

Rewrite $C_1$ as

$$C_1: (x+2)^2 + (y-8)^2 = 45$$

The line from the center of $C_1$, which is $(-2,8)$, to the point $(-5,14)$ is given by

$$L_1: y=-2x+4$$

So the tangent to $C_1$ at $(-5,14)$ is perpendicular to $L_1$ (so it has gradient $1/2$), and since it must pass through $(-5,14)$, it is given by

$$L_2: y = \frac 12 x+\frac {33}{2}$$

Rewrite circle $C_2$ as

$$C_2: (x+28)^2+(y+5)^2 = 65$$

Solve the two equations $L_2$ and $C_2$ simultaneously to obtain the intersection points

$$A = (-27,3) \qquad B = (-35, -1)$$

The center of $C_2$ is $O_2 = (-28,-5)$, and the lines from $O_2$ to each of $A$ and $B$ are respectively given by

$$O_2A : y=8x+219 \qquad O_2B: y = -\frac 47 x-21$$

So the tangents to $O_2$ at $A$ and $B$ have respective gradients $- 1/8$ and $7/4$, and are hence given, respectively, by

$$L_3: y = -\frac 18 x -\frac 38 \qquad L_4: y = \frac 74 x+ \frac{241}{4}$$

Finally, solve $L_3$ and $L_4$ simultaneously to obtain the intersection point being $$\biggl(-\frac {97}{3}, \frac {11}{3}\biggl)$$

Answer 2

This is easily solved without explicitly computing $A$, $B$ or the tangents at those points to $C_2$ by exploiting pole-polar relationships. The tangent to $C_1$ at $(-5,14)$ is the polar line of this point. The intersection of the tangents to $C_2$ at the points at which this line intersects $C_2$ is its pole. Thus, in homogeneous coordinates, $$C = \begin{bmatrix}1&0&28\\0&1&5\\28&5&744\end{bmatrix}^{-1} \begin{bmatrix}1&0&2\\0&1&-8\\2&-8&23\end{bmatrix} \begin{bmatrix}-5\\14\\1\end{bmatrix} = \frac1{13}\begin{bmatrix}-291\\33\\9\end{bmatrix}.$$ Dehomogenizing yields $\left(-\frac{97}3,\frac{11}3\right)$.