Question

Find the equation of the circle with:
(i) Centre (−2, 3) and radius 4.
(ii) Centre (a, b) and radius $\sqrt{a^2+b^2}$.
(iii) Centre (0, −1) and radius 1.
(iv) Centre (a cos α, a sin α) and radius a.
(v) Centre (a, a) and radius $\sqrt{2}$ a.

Answer 1

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

(i) Here, h = −2, k = 3 and a = 4

∴ Required equation of the circle:
${\left(x+2\right)}^2+{\left(y-3\right)}^2=4^2$
$\Rightarrow {\left(x+2\right)}^2+{\left(y-3\right)}^2=16$

(ii) Here, h = a, k = b and radius = $\sqrt{a^2+b^2}$

∴ Required equation of the circle:
${\left(x-a\right)}^2+{\left(y-b\right)}^2=a^2+b^2$
$\Rightarrow x^2+y^2-2ax-2by=0$

(iii) Here, h = 0, k = −1 and radius = 1

∴ Required equation of the circle:
${\left(x-0\right)}^2+{\left(y+1\right)}^2={\left(1\right)}^2$
$\Rightarrow x^2+y^2+2y=0$

(iv) Here, h = $a\cos \alpha$, k = $a\sin \alpha$ and radius = a

∴ Required equation of the circle:
${\left(x-a\cos \alpha \right)}^2+{\left(y-a\sin \alpha \right)}^2={\left(a\right)}^2$
$$\begin{gathered} \Rightarrow x^2+a^2{\cos }^2\alpha -2ax\cos \alpha +y^2+a^2{\sin }^2\alpha -2ay\sin \alpha =a^2 \\ \Rightarrow x^2+a^2\left({\sin }^2\alpha +{\cos }^2\alpha \right)-2ax\cos \alpha +y^2-2ay\sin \alpha =a^2 \\ \Rightarrow x^2+a^2-2ax\cos \alpha +y^2-2ay\sin \alpha =a^2 \\ \Rightarrow x^2+y^2-2ax\cos \alpha -2ay\sin \alpha =0 \end{gathered}$$

(v) Here, h = a, k = a and radius = $\sqrt{2}a$

∴ Required equation of the circle:
${\left(x-a\right)}^2+{\left(y-a\right)}^2={\left(\sqrt{2}a\right)}^2$
$$\begin{gathered} \Rightarrow x^2+a^2-2ax+y^2+a^2-2ay=2a^2 \\ \Rightarrow x^2+y^2-2ay-2ax=0 \end{gathered}$$