Which circle among the following bisects the circumference of the circle $x^2+y^2-8 x-6 y+23=0$ ?
Which circle among the following bisects the circumference of the circle $x^2+y^2-8 x-6 y+23=0$ ?
$x^2+y^2-6 x-4 y+9=0$
$x^2+y^2+6 x+4 y-9=0$
$x^2+y^2-6 x+4 y-9=0$
$x^2+y^2+6 x-4 y+9=0$
Solution
Given circle,
$
x^2+y^2-8 x-6 y+23=0
$
Centre of this circle is $(4,3)$.
If the equation of radical axis of circle $S_1$ and $S_2$ passes through $(4,3)$, then $S_2$ bisect the circumference of $S_1$.
Option (a) Let $S_2=x^2+y^2-6 x-4 y+9=0$
Then, equation of radical axis $(L)$ is
$
\begin{aligned}
L: S_1-S_2 & =0 \\
L:\left(x^2+y^2-8 x-6 y+23\right) & - \\
\left(x^2+y^2-6 x-4 y+9\right) & =0 \\
L:-2 x-2 y+14 & =0 \Rightarrow L: x+y-7=0
\end{aligned}
$
Check whether $(4,3)$ satisfy equation $L$.
LHS (4) + (3) $-7=0=$ RHS
$\therefore x^2+y^2-6 x-4 y+9=0$ bisect the
circumference of circle $x^2+y^2-8 x-6 y+23=0$
Similarly one can check for other options