The locus of the centres of the circles,which | vedclass

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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$.This circle cuts the given circles orthogonally.The condition for two circles $x

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$9x-10y+7=0$

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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$.This circle cuts the given circles orthogonally.The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally is $2(g_1g_2+f_1f_2) = c_1+c_2$.For the first circle $x^2+y^2+4x-6y+9=0$,we have $g_1=2, f_1=-3, c_1=9$. The condition gives $2(2g-3f) = c+9$,so $4g-6f-c=9$ ... $(ii)$.For the second circle $x^2+y^2-5x+4y+2=0$,we have $g_2=-2.5, f_2=2, c_2=2$. The condition gives $2(-2.5g+2f) = c+2$,so $-5g+4f-c=2$ ... $(iii)$.Subtracting $(iii)$ from $(ii)$,we get $(4g-6f-c) - (-5g+4f-c) = 9-2$,which simplifies to $9g-10f=7$.Replacing $(g, f)$ with $(x, y)$,the locus of the centre is $9x-10y=7$,or $9x-10y-7=0$.

The locus of the centres of the circles,which cut the circles $x^2+y^2+4x-6y+9=0$ and $x^2+y^2-5x+4y+2=0$ orthogonally,is

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