Two circles $k_1$ and $k_2$ with radii $r_1$ and $r_2$ have no common points. The line$ AB$ is a common internal tangent, and the line $CD$ is a common external tangent to these circles, where $A, C \in k_1$ and $B, D \in k_2$. Knowing that $AB=12$ and $CD =16$, find the value of the product $r_1r_2$.
Problem
Source:
Tags: geometry, Tangents, circles
natmath
15.02.2021 19:14
WLOG $r_2\geq r_1$ If $d$ is the distance between the centers of the circles, then the length of $AB$ and $CD$ respectively are $$\sqrt{d^2-(r_1+r_2)^2}=12$$$$\sqrt{d^2-(r_2-r_1)^2}=16$$ $$d^2-r_1^2-r_2^2-2r_1r_2=144$$$$d^2-r_1^2-r_2^2+2r_1r_2=256$$$$4r_1r_2=112$$$$r_1r_2=\boxed{28}$$
Euclides
16.02.2021 01:53
Let O1, O2 the two centers, and E the intersection of the two tangents. Let EC=AE=x (The two tangents from an exterior point to a circle are congruent). Also, therefore, DE=EB, that is 16 - x=12 + x; hence x = 2. Now, it is plain that triangles EAO1 and O2BE are similar: both have right angles at a point of tangency, and the acute angles of the one equal those of the other for, EO1 and EO2 being angle bisectors of the adjacent supplementary angles CEA and DEB, they are perpendicular and hence O1EA and O2EB are complementary, but O1EA and EO1A are complementary, hence angles O2EB and EO1A are congruent. Therefore the two said triangles are similar, having two angles of the one equal to two angles of the other, each to each. It follows that: O2B:AE = BE:AO1, that is r2:2 =(12 + 2):r1, or r1xr2 = 28.