Prove that 3 x^2+3 y^2-6 x+4 y-1=0, represents a circle. Find its… - Vidyadip

Question

Prove that $3 x^2+3 y^2-6 x+4 y-1=0$, represents a circle. Find its centre and radius.

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Answer

Given equation is
$
3 x^2+3 y^2-6 x+4 y-1=0
$
dividing by 3 , we get
$
x^2+y^2-2 x+\frac{4 y}{3}-\frac{1}{3}=0 \text { comparing }
$
this with
$
x^2+y^2+2 g x+2 f y+c=0
$
we get $2 g=-2$
$\therefore g=-1$
$2 f=\frac{4}{3} \quad \therefore f=\frac{2}{3}$ and $c=\frac{-1}{3}$
$g^2+f^2-c=(-1)^2+\left(\frac{2}{3}\right)^2-\frac{-1}{3}$
$=1+\frac{4}{9}+\frac{1}{3}=\frac{16}{9}$
As $\frac{16}{9}>0 \quad g^2+f-c>0$
$\therefore 3 x^2+3 y^2-6 x+4 y-1=0$ represents a circle.
$\therefore$ Centre of the circle $=(-g,-f)=\left(1, \frac{-2}{3}\right)$
Radius of circle $\mathrm{r}=\sqrt{(-g)^2+(-f)^2-c}$
$=\sqrt{\frac{16}{9}}=\frac{4}{3}$

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