Question

Find the coordinates of the centre and radius of each of the following circles :

(i) $x^2+y^2+6x-8y-24=0$
(ii) $2x^2+2y^2-3x+5y=7$
(iii) $\frac{1}{2}(x^2+y^2)+xcos\theta +ysin\theta -4=0.$
(iv) $x^2+y^2-ax-by=0$

Answer 1

The general equation of circles is
$x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
centre = (-g, -f)
radius $=\sqrt{g^2+f^2-c}\dots \left(A\right)$

(i) $x^2+y^2+6x-8y-24=0$
Hence $g=3,f=-4,c=-24$
Thus,
centre = (-3, 4)
radius $=\sqrt{g^2+f^2-c}=\sqrt{9+16+24}=\sqrt{49}$
$\therefore$ radius = 7

(ii) $2x^2+2y^2-3x+5y-7=0\Rightarrow x^2+y^2-\frac{3}{2}x+\frac{5}{2}y-\frac{7}{2}=0$
Hence, $g=\frac{-3}{4},f=\frac{5}{4},c=\frac{-7}{2}$
Thus,
Centre $=(\frac{3}{4}-\frac{5}{4})$
radius $=\sqrt{{\left(\frac{3}{4}\right)}^{+}{\left(\frac{5}{4}\right)}^2+\frac{7}{2}}=\sqrt{\frac{9}{16}+\frac{25}{16}+\frac{7}{2}}=\frac{\sqrt{90}}{4}$
$\therefore$ radius $=\frac{3\sqrt{10}}{4}$

(iii) $\frac{1}{2}(x^2+y^2)+xcos\theta +ysin\theta +-4=0\Rightarrow x^2+y^2+2xcos\theta +2ysin\theta -8=0$
Comparing with(i)
Hence $g=cos\theta ,f=sin\theta ,c=-8$
Thus.
centre =$(-cos\theta ,-sin\theta )$
radius =$\sqrt{co{s}^2\theta +si{n}^2\theta +8}=\sqrt{1+8}=3$
radius =3

(iv) $x^2+y^2-ax-by=0$
Hence $g=\frac{a}{2},f=-\frac{b}{2},c=0$
Thus,
centre $(=\frac{a}{2},\frac{b}{2})$
radius = $\sqrt{\frac{a^2}{4}+\frac{b^2}{4}+0}=\frac{\sqrt{a^2+b^2}}{2}$
radius $=\frac{\sqrt{a^2+b^2}}{2}$