If one of the diameters of the circle $x^2+y^2-10x+4y+13=0$ is a chord of another circle $C$, whose center is the point of intersection of the lines $2x+3y=12$ and $3x-2y=5$, then the radius of the circle $C$ is
Step-by-Step Solution
Key Concept: Centre of given circle: $(5,-2)$, radius $=\sqrt{25+4-13}=4$. Intersection of lines: $x=3$, $y=2$ — this is centre of circle $C$. Distance from $C$'s centre $(3,2)$ to given circle's centre $(5,-2)$: $CM=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$. The chord (diameter of given circle, length 8) has half-length 4. Radius of $C$: $\sqrt{CM^2+4^2}=\sqrt{20+16}=6$.
Centre of $C$: $(3,2)$. Given circle centre: $(5,-2)$, radius $4$. $CM=\sqrt{(5-3)^2+(-2-2)^2}=\sqrt{20}$. $CP=\sqrt{CM^2+(\text{half-chord})^2}=\sqrt{20+16}=6$.
Correct Answer: 3