# Finding the Angle Formed by the Common External Tangent Lines of Two Circles

The question is as follows:

The circle $x^2-6x+y^2-8y=24$ is transformed through a dilation into $x^2-14x+y^2-4y=-44$, point by point. The circles have two common external tangent lines, which meet at the dilation center. Find the size of the angle formed by these lines, and write an equation for each line.

I used Desmos to create the two circles. The center and radius for the original circle is $(3, 4)$ and $1$, respectively. The center and radius for the image circle is $(7,2)$ and $3$, respectively.

I know that one of the lines' equation has to be $y=5$. But I am not sure about the other one. Also, I have no idea about how I can calculate the angle created by the external tangent lines.

Any help will be greatly appreciated!

I would start by finding the position where the two tangents intersect. They also intersect with the line passing through the center of the circles. Suppose that the distance between centers is $d$, and the distance from the center of the small circle to the intersection is $x$. The two triangles formed by the lines through the centers, a tangent, and the corresponding radii are similar. You have then$$\frac{x}{d+x}=\frac{R_1}{R_2}$$ You can find $x$ from this equation. If $\alpha$ is the angle between tangents, the angle between one tangent and the line through the centers is $\alpha/2$. In the same figure, you can see that $$\sin\frac{\alpha}{2}=\frac{R_1}{x}$$ To find the equation of the other tangent, just note that is symmetric with respect to the line between centers. Transform one point from the first tangent by drawing the perpendicular to the line between centers, move twice the distance. Then you have the intersection point. You can therefore get the equation of the line