Question

Find the centre and radius of each of the following circles:
(i) (x − 1)² + y² = 4
(ii) (x + 5)² + (y + 1)² = 9
(iii) x² + y² − 4x + 6y = 5
(iv) x² + y² − x + 2y − 3 = 0.

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) Given:
(x − 1)² + y² = 4

Here, h = 1, k = 0 and a = 2

Thus, the centre is (1, 0) and the radius is 2.

(ii) Given:
(x + 5)² + (y + 1)² = 9

Here, h = −5, k = −1 and radius = 3

Thus, the centre is (−5, −1) and the radius is 3.

(iii) Given:
$x^2+y^2-4x+6y=5$

The given equation can be rewritten as follows:
${\left(x-2\right)}^2+{\left(y+3\right)}^2-4-9=5$
$\Rightarrow {\left(x-2\right)}^2+{\left(y+3\right)}^2=18$

Thus, the centre is (2, −3).
And, radius = $\sqrt{18}=3\sqrt{2}$

(iv) Given:
$x^2+y^2-x+2y-3=0$

The given equation can be rewritten as follows:
${\left(x-\frac{1}{2}\right)}^2+{\left(y+1\right)}^2-\frac{1}{4}-1-3=0$
$\Rightarrow {\left(x-\frac{1}{2}\right)}^2+{\left(y+1\right)}^2=\frac{17}{4}$

Thus, the centre is $\left(\frac{1}{2},-1\right)$ and and the radius is $\frac{\sqrt{17}}{2}$.