Question

Find the equations of the circle described on the common chord of the circle $x^2+y^2-4x5=0$ and $x^2+y^2+8y+7=0$ as diameter?

Answer 1

$Let,S_1=x^2+y^2-4x+5=0and,S_2=x^2+y^2+8y+7=0$

The equation of common chord is;

$S_1-S_2=0\Rightarrow (x^2+y^2-4x+5=0)-(x^2+y^2+8y+7=0)=0\Rightarrow 4x+8y+2=0$

Also the equation of circle passing through two circles is;

$S_1+\lambda S_2=0\Rightarrow (x^2+y^2-4x+5=0)+\lambda (x^2+y^2+8y+7=0)=0\Rightarrow (1+\lambda )x^2+(1+\lambda )y^2-4x+8y\lambda +(5+\lambda 7)=0\Rightarrow x^2+y^2-\frac{4x}{1+\lambda }+\frac{8y\lambda }{1+\lambda }+\frac{5+7\lambda }{1+\lambda }=0$

This is similar to equation of circle;

$x^2+y^2+2gx+2fy+c=0$

where centre is $(-g,-f)$

now finding value of $g$ and $f$

$g=\frac{2}{1+\lambda }andf=\frac{-4\lambda }{1+\lambda }$

hence centre of circle is $(\frac{2}{1+\lambda },\frac{-4\lambda }{1+\lambda })$

At this point lie on the common chord above $4x+8y+2=0$

it will satisfy the equation

putting the value in $4x+8y+2=0$

$\Rightarrow 4.\frac{2}{1+\lambda }-8.\frac{4\lambda }{1+\lambda }+2=0\Rightarrow \lambda =\frac{1}{3}$

Hence equation of circle is;

$\Rightarrow (1+\lambda )x^2+(1+\lambda )y^2-4x+8y\lambda +(5+\lambda 7)=0\Rightarrow (1+\frac{1}{3})x^2+(1+\frac{1}{3})y^2-4x+8y.\frac{1}{3}+(5+\frac{1}{3}.7)=0\Rightarrow 2x^2+2y^2-6x+4y+11=0$