Question

Find the coordinates of the centre and radius of each of the following circles:
(i) x² + y² + 6x − 8y − 24 = 0
(ii) 2x² + 2y² − 3x + 5y = 7
(iii) 1/2 (x² + y²) + x cos θ + y sin θ − 4 = 0
(iv) x² + y² − ax − by = 0

Answer 1

(i) The given equation can be rewritten as $x^2+y^2+2\left(3\right)x-2\left(4\right)y-24=0$.

∴ Centre = $\left(-3,-4\right)$
And, radius = $\sqrt{{\left(3\right)}^2+{\left(4\right)}^2+24}=\sqrt{49}=7$


(ii) The given equation can be rewritten as $x^2+y^2-\frac{3x}{2}+\frac{5y}{2}-\frac{7}{2}=0$.

∴ Centre = $\left(\frac{3}{4},\frac{-5}{4}\right)$
And, radius = $\sqrt{{\left(\frac{3}{4}\right)}^2+{\left(\frac{-5}{4}\right)}^2+\frac{7}{2}}=\sqrt{\frac{34+56}{16}}=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}{4}$


(iii) The given equation can be rewritten as $x^2+y^2+2x\cos \theta +2y\sin \theta -8=0$.

∴ Centre = $\left(-\cos \theta ,-\sin \theta \right)$
And, radius = $\sqrt{{\left(-\cos \theta \right)}^2+{\left(-\sin \theta \right)}^2+8}=\sqrt{1+8}=3$


(iv) The given equation can be rewritten as $x^2+y^2-\frac{2ax}{2}-\frac{2by}{2}=0$.

∴ Centre = $\left(\frac{a}{2},\frac{b}{2}\right)$
And, radius = $\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$