Question

Find the equation of a circle which is coaxial with the circles $2x^2+2y^2-2x+6y-3=0$ and $x^2+y^2+4x+2y+1=0$.
It is given that the centre of the circle to be determined lies on the radical axis of these circles.

Answer 1

The two given circles in standard form we
$S_1=0$ or $x^2+y^2-x+3y-\frac{3}{2}=0$
$S_2=0$ or $x^2+y^2+4x+2y+1=0$
Radical axis is given by $S_1-S_2=0$.
$\therefore -5x+y-\frac{5}{2}=0$
or $10x-2y+5=0.....\left(1\right)$
Circle coaxial with the given circles is
$S_1+\lambda S_2=0$
$(x^2+y^2-x+3y-\frac{3}{2})+\lambda (x^2+y^2+4x+2y+1)=0$
or $(x^2+y^2)(1+\lambda )-(1-4\lambda )x+(3+2\lambda )y-\frac{3}{2}+\lambda =0$
or $x^2+y^2-\frac{1-4\lambda }{1+\lambda }x+\frac{3+2\lambda }{1+\lambda }y+\frac{\lambda -\frac{3}{2}}{1+\lambda }=0$,
Its centre $(\frac{1}{2}\cdot \frac{1-4\lambda }{1+\lambda },-\frac{1}{2}\cdot \frac{3+2\lambda }{1+\lambda })$
i.e. $(-g,-f)$ lies on $\left(1\right)$
$\therefore \frac{5(1-4\lambda )}{1+\lambda }+\frac{3+2\lambda }{1+\lambda }+5=0$
or $5-20\lambda +3+2\lambda +5+5\lambda =0$
or $13-13\lambda =0\therefore \lambda =1$
Putting in $\left(2\right)$, the required circle is
$x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{1}{4}=0$
or $4x^2+4y^2+6x+10y-1=0$.