Let x-4=0 be the radical axis of two circles | vedclass

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(B) The equation of the first circle is $x^2+y^2-36=0$. Let the equation of the second circle be $x^2+y^2+2gx+2fy+c=0$. The radical axis of two circle

Vedclass · Accepted Answer

$x^2+y^2-18x+36=0$

Vedclass · Answer

(B) The equation of the first circle is $x^2+y^2-36=0$. Let the equation of the second circle be $x^2+y^2+2gx+2fy+c=0$. The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. Given the radical axis is $x-4=0$,we can write the second circle as $x^2+y^2-36+k(x-4)=0$,which simplifies to $x^2+y^2+kx-4k-36=0$. 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 intersect orthogonally,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$. Here,$g_1=0, f_1=0, c_1=-36$ and $g_2=k/2, f_2=0, c_2=-4k-36$. Substituting these into the condition: $2(0)(k/2) + 2(0)(0) = -36 + (-4k-36)$. $0 = -72 - 4k$ $\Rightarrow 4k = -72$ $\Rightarrow k = -18$. Substituting $k=-18$ into the equation of the second circle: $x^2+y^2-18x-4(-18)-36=0$. $x^2+y^2-18x+72-36=0 \Rightarrow x^2+y^2-18x+36=0$.

Let $x-4=0$ be the radical axis of two circles which are intersecting orthogonally. If $x^2+y^2=36$ is one of those circles,then the other circle is

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