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(A) The given circle is $x^{2}+y^{2}+4x+6y-12=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$, we get $g=2, f=3, c=-12$. The centre $C_{1}$ is $(-g, -f)

Vedclass · Accepted Answer

$\sqrt{41} 	ext{ unit}$

Vedclass · Answer

(A) The given circle is $x^{2}+y^{2}+4x+6y-12=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$, we get $g=2, f=3, c=-12$. The centre $C_{1}$ is $(-g, -f) = (-2, -3)$ and the radius $r_{1} = \sqrt{g^{2}+f^{2}-c} = \sqrt{2^{2}+3^{2}-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5 	ext{ units}$. Since a diameter of this circle is a chord of circle $S$, the distance $d$ from the centre of $S$, $C_{2}(2, -3)$, to this chord is $0$ (because the chord passes through the centre of the first circle, but the problem states the diameter is a chord of $S$, implying the chord is a diameter of the first circle). Actually, the chord is a diameter of the first circle, so its length is $2r_{1} = 10$. The distance $d$ from $C_{2}(2, -3)$ to the centre $C_{1}(-2, -3)$ is $\sqrt{(2 - (-2))^{2} + (-3 - (-3))^{2}} = \sqrt{4^{2} + 0^{2}} = 4$. Let $R$ be the radius of circle $S$. The distance from the centre $C_{2}$ to the chord is $d=4$. The half-length of the chord is $a = r_{1} = 5$. Using the Pythagorean theorem in the triangle formed by the radius $R$, distance $d$, and half-chord $a$: $R^{2} = d^{2} + a^{2} = 4^{2} + 5^{2} = 16 + 25 = 41$. Therefore, $R = \sqrt{41} 	ext{ units}$.

If one of the diameters of the circle given by the equation $x^{2}+y^{2}+4x+6y-12=0$ is a chord of a circle $S$ whose centre is $(2,-3)$, then the radius of $S$ is:

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