The length of the diameter of the circle whic | vedclass

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(C) Let the equation of the required circle be $x^{2}+y^{2}+2gx+2fy+c=0$. Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2

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$4$

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(C) Let the equation of the required circle be $x^{2}+y^{2}+2gx+2fy+c=0$. Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ cut orthogonally if $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$. For the given circles,the conditions are: $1) -g-f = c-14 \implies -g-f-c = -14$ $2) 3g-5f = c-10 \implies 3g-5f-c = -10$ $3) -2g+3f = c-27 \implies -2g+3f-c = -27$ Subtracting $(1)$ from $(2)$: $4g-4f = 4 \implies g-f = 1 \implies g = f+1$. Subtracting $(1)$ from $(3)$: $-g+4f = -13$. Substituting $g = f+1$: $-(f+1)+4f = -13 \implies 3f = -12 \implies f = -4$. Then $g = -4+1 = -3$. From $(1)$: $-(-3)-(-4) = c-14 \implies 3+4 = c-14 \implies c = 21$. The circle is $x^{2}+y^{2}-6x-8y+21=0$. The radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-3)^{2}+(-4)^{2}-21} = \sqrt{9+16-21} = \sqrt{4} = 2$. The diameter is $2r = 2 	imes 2 = 4$.

The length of the diameter of the circle which cuts the following three circles orthogonally is:
$x^{2}+y^{2}-x-y-14=0$
$x^{2}+y^{2}+3x-5y-10=0$
$x^{2}+y^{2}-2x+3y-27=0$

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The length of the diameter of the circle which cuts the following three circles orthogonally is:$x^{2}+y^{2}-x-y-14=0$$x^{2}+y^{2}+3x-5y-10=0$$x^{2}+y^{2}-2x+3y-27=0$