Common chord of two intersecting circles

Text

Two circles have radii $15$ cm and $20$ cm and their centers are $25$ cm apart. Find the length of the common chord.

Solution

Let $x$ be the distance from the first circle’s center to the foot of the common chord, measured along the line of centers ($d=25$): $x=\frac{d^2+r_1^2-r_2^2}{2d}=\frac{625+225-400}{50}=\frac{450}{50}=9.$ The half-chord is $h=\sqrt{r_1^2-x^2}=\sqrt{225-81}=\sqrt{144}=12,$ so the common chord is $\overline{AB}=2h=24$ cm. $\boxed{\overline{AB}=24\ \text{cm}}$