The circle S=0 cuts the circle x^2+y^2-4x+2y- | vedclass

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(A) Given that,the circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally and the centre of the circle $S=0$ is $(2,3)$.Let the equation of the

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$2$

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(A) Given that,the circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally and the centre of the circle $S=0$ is $(2,3)$.Let the equation of the circle $S=0$ be $x^2+y^2+2gx+2fy+c=0$. Since the centre is $(-g, -f) = (2,3)$,we have $g=-2$ and $f=-3$.As we know that,if two circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g'x+2f'y+c'=0$ intersect orthogonally,then $2gg'+2ff'=c+c'$.Here,$(g, f) = (-2, -3)$ and for the given circle $x^2+y^2-4x+2y-7=0$,$(g', f') = (-2, 1)$ and $c' = -7$.Substituting these values into the condition:$2(-2)(-2) + 2(-3)(1) = c + (-7)$$8 - 6 = c - 7$$2 = c - 7$$c = 9$The radius $r$ of the circle $S=0$ is given by $\sqrt{g^2+f^2-c}$.$r = \sqrt{(-2)^2 + (-3)^2 - 9} = \sqrt{4 + 9 - 9} = \sqrt{4} = 2$.

The circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally. If $(2,3)$ is the centre of the circle $S=0$,then its radius is

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