Find the centre and radius of each of the following circles.
(i) $(x - 1)^{2} + y^{2} = 4$
(ii) $(x + 5)^{2} + (y + 1)^{2} = 9$
(iii) $(x^{+} y^{2} - 4 x + 2 y - 3 = 0$

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Solution

(i) The general equation of circle is
$(x - a)^{2} + (y — b)^{2} r^{2}$
or $x^{2} + y^{2} — 2 a x — 2 b y + a^{2} + b^{2} = r^{1} \dots (A)$
Where (a, b) is the centre and r is radius of the circle.

(i) $(x - 1)^{2} + y^{2} == 4 \Rightarrow (x - 1)^{2} + (y - 0)^{2} = 2^{2}$
Comparing with (A) we get,
(1, 0) is the centre 2 is the radius

(ii) $(x + 5)^{2} + (y + 1)^{2} = 9 \Rightarrow (x + 5)^{2} + (y + 1)^{2} = 3^{2} [x - (- 5)]^{+} [y - (- 1)]^{=} (3)^{2}$
Comparing with (A) we get,
centre = (-5, -1)
radius =3

(ii) $x^{2} + y^{2} — 4 x + 6 y = 5 \Rightarrow (x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) = 5 + 4 + 9$
(Add 4 and 9 on both sides)
$(x - 2)^{2} + (y + 3)^{2} = 18 (x - 2)^{2} + (y + 3)^{2} = 3 \sqrt{2}$
Comparing with (A) we get,
centre = (2, -3)
radius = $3 \sqrt{2}$

(iv) $x^{2} + y^{2} - x + 2 y = 3$
Add $\frac{1}{4}$ and 1 on both sides
$\Rightarrow (x^{2} - x + \frac{1}{4}) + (y^{2} + 2 y + 1) = 3 + \frac{1}{4} + 1 \Rightarrow {(x - \frac{1}{2})}^{2} + (y + 1)^{2} = {(\frac{\sqrt{17}}{2})}^{2}$
Comparing with (A) we get,
centre $= (\frac{1}{2}, - 1)$
radius= $\frac{\sqrt{17}}{2}$

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