If one of the diameters of the circle x^2+y^2-10x+4y+13=0 is a chord of another circle C, …

Question

If one of the diameters of the circle $x^2+y^2-10x+4y+13=0$ is a chord of another circle $C$, whose centre is the point of intersection of the lines $2x+3y=12$ and $3x-2y=5$, then the radius of the circle $C$ is:

Accepted Answer

Given circle: centre $(5,-2)$, radius $=\sqrt{25+4-13}=4$.

Centre of $C$: solving $2x+3y=12,\ 3x-2y=5$ gives $(3,2)$.

The diameter of the given circle (length $8$, midpoint $(5,-2)$) is a chord of $C$:

$$d=\sqrt{(5-3)^2+(-2-2)^2}=\sqrt{20}.$$

$$R^2=d^2+4^2=20+16=36 \;\Rightarrow\; R=6.$$

Solutions · 1