Question

Find the centre and radius of the circle :
$(x+2)(x-1)+(y-1)(y+3)=0$

Answer 1

The equation of the circle is
$(x+2)(x-1)+(y-1)(y-3)=0$ .......(i)
$\Rightarrow$ $\{x-(-2\left)\right\}(x-1)+(y-1)\{y-(-3\left)\right\}=0$
It shows that the end-point of one diameter of circle (i) are $(-2,1)$ and$(1,-3)$.
$\therefore$ Centre$=(\frac{-2+1}{2},\frac{1-3}{2})=(-\frac{1}{2},-1)$
and Radius$=$ Distance between $(-\frac{1}{2},-1)$ and $(-2,1)$
$=\sqrt{{(-2+\frac{1}{2})}^2+{(1+1)}^2}$
$=\sqrt{\frac{9}{4}+4}=\frac{5}{2}$

Thus, centre and radius is $(-\frac{1}{2},-1)$ and $\frac{5}{2}$.