A system of Circles pass through $(2,3)$ and have their centers on the line $x+2y-7=0$.
Show that the chords in which the circles of the given system intersects the circle $S_1:x^2+y^2-8x+6y-9=0$ are concurrent and also find the point of concurrency.
ATTEMPT:
The circles of the given system must intersect the line $x+2y-7=0.$
Therefore the equation of family of circles passing through $(2,3)$ and $(7-2k,k)$ , for any arbitrary $k$ can be written as $S$: $(x-2)(x-7+2k) +(y-3)(y-k) +c $$\begin{vmatrix} 1 & x & y \\ 1 & 7-2k & k \\ 1 & 2 & 3 \\ \end{vmatrix} $$ =0$ ,where $c$ is any arbitrary constant.
Now using the condition that the center lies on the line $x+2y-7=0$ we can get a relation in $k$ and $c.$
And the equation of common chords can be given by $S-S_1=0.$
But how to prove that the chords are concurrent?
The center of the system of circle lies on $x+2y-7=0$, if the center is $(h,k)$ then $h+2k-7=0$
And as the circles passes through $(2,3)$, therefore the radius is $\sqrt{(h-2)^2+(k-3)^2}$
Therefore, the equation of the circle is
$(x-h)^2+(y-k)^2=(h-2)^2+(k-3)^2$ i.e., $x^2+y^2-2xh-2yh+4h+6k+13=0$ we call this $S$
And the fixed circle is $x^2+y^2-8x+6y-9=0$----$S-1$
The common chord of $S$ and $S_1$ is $S-S_1=0$
i.e., $-8x+6y-9+2xh+2yk-4h-6k-13=0$
or, $-8x+6y-9+2xh+(7-h)y-4h-3(7-h)-13=0$
or, $-8x+6y+7y-9-21-13+h(2x-y-4-3)$
or, $-8x+13y-43+h(2x-y-7)=0$
This is a system of lines which always passes through through the intersection of $-8x+13y-43=0$ and $(2x-y-7)=0$
Therefore, the common chords are concurrent.