Question

Find the centre and radius of circle
(a)$x^2+y^2+8x+10y-8=0$
(b)$x^2+y^2-16x+20y-24=0$
(c)$3x^2+3y^2-x=0$
(d)$5x^2+5y^2-y=0$

Answer 1

$\begin{array}{c}\left(a\right)x^2+y^2+8x+10y-8=0\\ \Rightarrow centre=\left(-g,-f\right)=\left(-4,-5\right)\\ \Rightarrow Radius=\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2-\left(-8\right)}\\ =\sqrt{16+25+8}\\ \Rightarrow Radius=\sqrt{49}unit\\ Radius=7unit\\ \left(b\right)x^2+y^2-16x+20y-24=0\\ \Rightarrow centre=\left(-g,-f\right)=\left(8,-10\right)\\ \Rightarrow Radius=\sqrt{{\left(8\right)}^2+{\left(-10\right)}^2-\left(-24\right)}\\ =\sqrt{64+100+24}\\ =\sqrt{188}=2\sqrt{47}unit\\ \left(c\right)3x^2+3y^2-x=0\\ \Rightarrow x^2+y^2-\frac{x}{3}=0\\ \Rightarrow centre=\left(-g,-f\right)=\left(\frac{1}{6},0\right)\\ \Rightarrow Radius=\sqrt{{\left(\frac{1}{6}\right)}^2-0}=\frac{1}{6}unit\\ \left(d\right)5x^2+5y^2-y=0\\ x^2+y^2-\frac{y}{5}=0\\ \Rightarrow centre=\left(0,\frac{1}{10}\right)\\ \Rightarrow Radius=\sqrt{{\left(\frac{1}{10}\right)}^2}\\ =\frac{1}{10}unit\end{array}$