If one of the diameters of the circle x^2+y^2 | vedclass

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(C) The given circle is $x^2+y^2-10x+4y+13=0$. Its center $M$ is $(5, -2)$ and its radius $r$ is $\sqrt{5^2+(-2)^2-13} = \sqrt{25+4-13} = \sqrt{16} =

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$6$

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(C) The given circle is $x^2+y^2-10x+4y+13=0$. Its center $M$ is $(5, -2)$ and its radius $r$ is $\sqrt{5^2+(-2)^2-13} = \sqrt{25+4-13} = \sqrt{16} = 4$.The center of circle $C$ is the intersection of $2x+3y=12$ and $3x-2y=5$. Solving these,we get $x=3$ and $y=2$. So,the center of $C$ is $O(3, 2)$.The diameter of the first circle is a chord of circle $C$. Let $P$ be a point on the circumference of circle $C$ such that $MP$ is the radius of the first circle,i.e.,$MP = 4$. Since $M$ is the center of the first circle and $MP$ is a radius,$M$ lies on the chord.The distance $d$ between the centers $O(3, 2)$ and $M(5, -2)$ is $d = \sqrt{(5-3)^2 + (-2-2)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4+16} = \sqrt{20}$.In the right-angled triangle formed by the center of $C$,the center of the first circle,and a point on the chord,the radius $R$ of circle $C$ is given by $R^2 = d^2 + r^2 = 20 + 16 = 36$. Thus,$R = 6$.

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

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