If x-4=0 is the radical axis of two orthogona | vedclass

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(B) Let the equation of the first circle be $S_1 \equiv x^2+y^2-36=0$. Let the equation of the second circle be $S_2 \equiv x^2+y^2+2gx+2fy+c=0$. The

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$(9,0)$

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(B) Let the equation of the first circle be $S_1 \equiv x^2+y^2-36=0$. Let the equation of the second circle be $S_2 \equiv x^2+y^2+2gx+2fy+c=0$. The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. Thus,$(x^2+y^2-36) - (x^2+y^2+2gx+2fy+c) = 0$,which simplifies to $-2gx-2fy-36-c=0$. Given the radical axis is $x-4=0$,we can write it as $x+0y-4=0$. Comparing the coefficients,we get $\frac{-2g}{1} = \frac{-2f}{0} = \frac{-36-c}{-4} = k$. This gives $f=0$ and $2g = -k$,so $g = -k/2$. Also,$36+c = 4k$,so $c = 4k-36$. Since the circles are orthogonal,the condition is $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1=0, f_1=0, c_1=-36$ and $g_2=g, f_2=f, c_2=c$. Substituting these,$2(0)(g) + 2(0)(f) = -36 + c$. So,$0 = -36 + c$,which means $c = 36$. From $c = 4k-36$,we have $36 = 4k-36$,so $4k = 72$,which gives $k = 18$. Then $g = -k/2 = -18/2 = -9$. The centre of the second circle is $(-g, -f) = (9, 0)$.

If $x-4=0$ is the radical axis of two orthogonal circles,out of which one is $x^2+y^2=36$,then the centre of the other circle is

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