Olympiad Maths Prep

Track / Stage 5 / 128 of 400 #728 of 2000

Problem 728

AIME late
Geometry Difficulty 5.4 Find the answer

Example 13 Find the equation of the circle that is tangent to the circle C:x2+y26x8y+17=0C: x^{2}+y^{2}-6 x-8 y+17=0 at the point (1,2)(1,2) and has a radius of 522\frac{5 \sqrt{2}}{2}.

Official solution

Consider the point (1,2)(1,2) as the point circle curve Γ0:(x1)2+(y2)2=0\Gamma_{0}:(x-1)^{2}+(y-2)^{2}=0. By property 9, the equation of the required circle is x2+y26x8y17+λ[(x1)2+(y2)2]=0x^{2}+y^{2}-6 x-8 y-17+\lambda\left[(x-1)^{2}+(y-2)^{2}\right]=0.
That is, (x3λ1+λ)2+(y4+2λ1+λ)2=8(1+λ)2\left(x-\frac{3 \lambda}{1+\lambda}\right)^{2}+\left(y-\frac{4+2 \lambda}{1+\lambda}\right)^{2}=\frac{8}{(1+\lambda)^{2}}.
From 8(1+λ)2=(522)2\frac{8}{(1+\lambda)^{2}}=\left(\frac{5 \sqrt{2}}{2}\right)^{2}, we get λ1=15\lambda_{1}=-\frac{1}{5} or λ2=95\lambda_{2}=-\frac{9}{5}.
Thus, we obtain C1:(x72)2+(y92)2=(522)2C_{1}:\left(x-\frac{7}{2}\right)^{2}+\left(y-\frac{9}{2}\right)^{2}=\left(\frac{5 \sqrt{2}}{2}\right)^{2} and C2:(x+32)2+(y+12)2=(522)2C_{2}:\left(x+\frac{3}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{5 \sqrt{2}}{2}\right)^{2}.
It is easy to see that the point (1,2)(1,2) satisfies the equation of circle C1C_{1}, and the distance between the centers of circle C1C_{1} and circle CC equals the absolute value of the difference of their radii, so circle C1C_{1} is internally tangent to circle CC at point (1,2)(1,2), and circle C1C_{1} is the required one.
Similarly, circle C1C_{1} is externally tangent to circle CC at point (1,2)(1,2), and circle C2C_{2} is also the required one.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.