If the circle x^{2}+y^{2}+2gx+2fy+c=0 cuts th | vedclass

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Question

(D) The condition that a circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts another circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$ at the extremities of its diameter

Vedclass · Accepted Answer

$4f=3g$

Vedclass · Answer

(D) The condition that a circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts another circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$ at the extremities of its diameter is equivalent to the condition of orthogonality: $2(gg_{i}+ff_{i})=c+c_{i}$.For the first circle $x^{2}+y^{2}-5=0$, $g_{1}=0, f_{1}=0, c_{1}=-5$. Applying the condition: $2(g(0)+f(0))=c-5 \Rightarrow c=5$.For the second circle $x^{2}+y^{2}-8x-6y+10=0$, $g_{2}=-4, f_{2}=-3, c_{2}=10$. Applying the condition: $2(g(-4)+f(-3))=c+10$ $\Rightarrow 2(-4g-3f)=5+10$ $\Rightarrow 4g+3f=-15/2$.For the third circle $x^{2}+y^{2}-4x+2y-2=0$, $g_{3}=-2, f_{3}=1, c_{3}=-2$. Applying the condition: $2(g(-2)+f(1))=c-2$ $\Rightarrow 2(-2g+f)=5-2$ $\Rightarrow -2g+f=3/2$.Solving the system of equations $4g+3f=-7.5$ and $-2g+f=1.5$:From the second equation, $f=2g+1.5$. Substituting into the first: $4g+3(2g+1.5)=-7.5$ $\Rightarrow 10g+4.5=-7.5$ $\Rightarrow 10g=-12$ $\Rightarrow g=-1.2$.Then $f=2(-1.2)+1.5 = -2.4+1.5 = -0.9$.Checking option $D$: $4f = 4(-0.9) = -3.6$ and $3g = 3(-1.2) = -3.6$. Thus, $4f=3g$ is correct.

If the circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts the three circles $x^{2}+y^{2}-5=0$, $x^{2}+y^{2}-8x-6y+10=0$, and $x^{2}+y^{2}-4x+2y-2=0$ at the extremities of their diameters, then

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