If a circle S passing through the point (3,4) | vedclass

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(D) Let the equation of the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. The centre of this circle is $(-g, -f)$.Since the circle passes through the point $(3

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$6x+8y+11=0$

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(D) Let the equation of the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. The centre of this circle is $(-g, -f)$.Since the circle passes through the point $(3,4)$,we have:$3^2+4^2+2g(3)+2f(4)+c=0$$9+16+6g+8f+c=0$$6g+8f+c+25=0$ ... $(i)$Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ intersect orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.Here,the given circle is $x^2+y^2-36=0$,so $g_2=0, f_2=0, c_2=-36$.Applying the condition for orthogonal intersection:$2g(0)+2f(0)=c-36$$c=c-36$ (This implies $c=-36$)Substituting $c=-36$ into equation $(i)$:$6g+8f-36+25=0$$6g+8f-11=0$Let the centre of the circle $S$ be $(x, y)$,where $x=-g$ and $y=-f$. Thus,$g=-x$ and $f=-y$.Substituting these into the equation $6g+8f-11=0$,we get:$6(-x)+8(-y)-11=0$$-6x-8y-11=0$$6x+8y+11=0$

If a circle $S$ passing through the point $(3,4)$ cuts the circle $x^2+y^2=36$ orthogonally,then the locus of the centre of $S$ is

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