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Question

(A) Let the required circle be $x^2+y^2+2gx+2fy+c=0 \dots(1)$.Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally

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$x^2+y^2-16x-18y-4=0$

Vedclass · Answer

(A) Let the required circle be $x^2+y^2+2gx+2fy+c=0 \dots(1)$.Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.Applying this condition to circle $(1)$ and the given circles:For circle $x^2+y^2-2x+3y-7=0$: $2g(-1)+2f(3/2)=c-7 \Rightarrow -2g+3f-c=-7 \dots(2)$.For circle $x^2+y^2+5x-5y+9=0$: $2g(5/2)+2f(-5/2)=c+9 \Rightarrow 5g-5f-c=9 \dots(3)$.For circle $x^2+y^2+7x-9y+29=0$: $2g(7/2)+2f(-9/2)=c+29 \Rightarrow 7g-9f-c=29 \dots(4)$.Subtracting $(2)$ from $(3)$: $(5g-5f-c) - (-2g+3f-c) = 9 - (-7) \Rightarrow 7g-8f=16 \dots(5)$.Subtracting $(3)$ from $(4)$: $(7g-9f-c) - (5g-5f-c) = 29 - 9$ $\Rightarrow 2g-4f=20$ $\Rightarrow g-2f=10$ $\Rightarrow g=2f+10$.Substituting $g$ into $(5)$: $7(2f+10)-8f=16$ $\Rightarrow 14f+70-8f=16$ $\Rightarrow 6f=-54$ $\Rightarrow f=-9$.Then $g=2(-9)+10=-8$.From $(2)$: $-2(-8)+3(-9)-c=-7$ $\Rightarrow 16-27-c=-7$ $\Rightarrow -11-c=-7$ $\Rightarrow c=-4$.Substituting $g, f, c$ in $(1)$: $x^2+y^2-16x-18y-4=0$.

Find the equation of the circle which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$,$x^2+y^2+5x-5y+9=0$,and $x^2+y^2+7x-9y+29=0$.

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