The distance between the centres of two circles is $d$. The lengths of the direct and transverse common tangents are $L$ and $M$ respectively. If $L^2 + M^2 = 320$ and the sum of the squares of the radii is 160, what is $d$?

Question

Accepted Answer

8\sqrt{5}. For two circles, the direct common tangent length satisfies $L^2=d^2-(r_1-r_2)^2$ and the transverse common tangent length satisfies $M^2=d^2-(r_1+r_2)^2$. Adding them gives $L^2+M^2=2d^2-2(r_1^2+r_2^2)$. Substituting $320=2d^2-2(160)$ gives $320=2d^2-320$, so $d^2=320$ and $d=8\sqrt{5}$.

The distance between the centres of two circles is $d$. The lengths of the direct and transverse common tangents are $L$ and $M$ respectively. If $L^2 + M^2 = 320$ and the sum of the squares of the radii is 160, what is $d$?

Solution

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