Length of direct common tangent of orthogonally intersecting circles?

Question

This is a question I found: Given two circles intersecting orthogonally having the length of common chord 24/5 units the radius of one of the circles is 3 units then what is the length of the direct common tangent? The answer is 2√6. How do I approach this problem? I have no idea what to do. Part of the common chord, the radius of one circle and another third side could form four right triangles, and the radius and common tangent with a third side could form one right triangle each by each circle, but I don't see how that helps. What should I do?

Answer 1

Sorry for hand-drawn picture.

Here $O_1$ is the center of the first circle (which is of radius 3), $O_2$ is the center of the second circle, $AB$ is the common chord, $UV$ is the common tangent. A point $E$ is the intersection of $AB$ and $O_1O_2$.

Let us at first find a radius of the second circle. Let $r$ be this radius. Consider a triangle $A O_1 O_2$. On one hand its are equals $3r/2$ (since circles are intersecting orthogonally). On the other hand its area equals to $\frac{1}{2} \cdot \frac{12}{5} \cdot \sqrt{9 + r^2}$ (since $AE$ is perpendicular to $O_1 O_2$ and its length is $12/5$). Thus we get

$$r = \frac{4}{5} \sqrt{9 + r^2} \implies r = 4.$$

Along the way we obtain that $|O_1 O_2| = 5$.

Draw a line parallel to $UV$ through $O_1$. Let $W$ be intersection of this line with $V O_2$. Observe that $UVWO_1$ is a rectangle. Thus $|VW| = |O_1U| = 3$ and $$|O_2 W| = |O_2 V| - |VW| = |O_2V| - |O_1U| = 4 - 3 = 1.$$

By noting that $O_1 W O_2$ is a right tringle, we get $$|UV| = |O_1W| = \sqrt{|O_1O_2|^2 - |O_2W|^2} = \sqrt{24} = 2\sqrt{6}.$$