Question

Prove that the length of the common chord of the circles $x^2+y^2+ax+by+c=0$ and $x^2+y^2+bx+ay+c=0$ is $\sqrt{\frac{1}{2}(a+b{)}^2-4c.}$

Answer 1

Here also common chord is x- y = 0
$C(-\frac{a}{2},-\frac{b}{2})andr=\frac{1}{2}\sqrt{a^2+b^2-4c}$
$p=\frac{(-a/2)-(-b/2)}{\sqrt{2}}=\frac{b-a}{2\sqrt{\left(2\right)}}$
Length $AB=2\sqrt{r^2-p^2}=\sqrt{4r^2-4p^2}$
$\therefore AB=\sqrt{(a^2+b^2-4c)-{\left(\frac{b-a}{\sqrt{2}}\right)}^2}$
$\sqrt{\frac{(a+b{)}^2-8c}{2}}$