Question

Find the centre and radius of the circle $x^2+y^2+3x+5y=0$

Answer 1

Given:$x^2+y^2+3x+5y=0$

$\Rightarrow x^2+2\times \frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+y^2+2\times \frac{5}{2}y+\frac{25}{4}-\frac{25}{4}+11=0$

$\Rightarrow {(x+\frac{3}{2})}^2+{(y+\frac{5}{2})}^2-\frac{9}{4}-\frac{25}{4}=0$

$\Rightarrow {(x+\frac{3}{2})}^2+{(y+\frac{5}{2})}^2-\frac{9+25}{4}=0$

$\Rightarrow {(x+\frac{3}{2})}^2+{(y+\frac{5}{2})}^2-\frac{34}{4}=0$

$\Rightarrow {(x+\frac{3}{2})}^2+{(y+\frac{5}{2})}^2-\frac{17}{2}=0$

$\Rightarrow {(x+\frac{3}{2})}^2+{(y+\frac{5}{2})}^2=\frac{17}{2}$

which is comparable with ${(x-h)}^2+{(y-k)}^2=r^2$

where $h=\frac{-3}{2},k=\frac{-5}{2}$ and $r=\sqrt{\frac{17}{2}}$

Hence, the given circle has centre $(\frac{-3}{2},\frac{-5}{2})$ and its radius$=\sqrt{\frac{17}{2}}$