Question

Prove that
$3x² + {3y}^{2} - 6x + 4y - 1 = 0$
represents a circle. Find its centre and radius.​

Answer 1

Basic Concept Used :-

In order to show that the equation represents a circle is it enough to get it into the form of

$\sf \: {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2}$

$\sf \: where \: (h, k) \: represents \: centre \: and \: r \: represent \: radius.$

Let's solve the problem now!!

Given equation is

$\rm :\longmapsto\: {3x}^{2} + {3y}^{2} - 6x + 4y - 1 = 0$

$\rm :\longmapsto\:( {3x}^{2} - 6x) + ( {3y}^{2} + 4y) - 1 = 0$

$\rm :\longmapsto\:3( {x}^{2} - 2x) + 3( {y}^{2} + \dfrac{4}{3}y ) = 1$

• Divide both sides by 3, we get

$\rm :\longmapsto\: {x}^{2} - 2x + {y}^{2} - \dfrac{4}{3}y = \dfrac{1}{3}$

• Now adding 1 and 4/9 on both sides

$\rm :\longmapsto\:( {x}^{2} - 2x + 1) + ( {y}^{2} + \dfrac{4y}{3} + \dfrac{4}{9}) = \dfrac{1}{3} + 1 + \dfrac{4}{9}$

$\rm :\longmapsto\: {(x - 1)}^{2} + {\bigg( y + \dfrac{2}{3} \bigg) }^{2} = \dfrac{3 + 9 + 4}{9}$

$\rm :\longmapsto\: {(x - 1)}^{2} + {\bigg( y + \dfrac{2}{3} \bigg) }^{2} = \dfrac{16}{9}$

$\rm :\longmapsto\: {(x - 1)}^{2} + {\bigg( y + \dfrac{2}{3} \bigg) }^{2} = {\bigg(\dfrac{4}{3} \bigg) }^{2}$

$\bf\implies \:its \: an \: equation \: of \: circle$

whose

$\rm :\longmapsto\:Centre \: is \: \bigg(1, - \: \dfrac{2}{3} \bigg)$

and

$\rm :\longmapsto\:Radius \: is \: \dfrac{4}{3}$

More Information :-

$\sf \: The \: general \: equation \: of \: circle \: is \:$

$\sf \: {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$

then

$\sf \: Centre \: = \: ( - \: g, \: - \: f)$

and

$\sf \: Radius \: = \: \sqrt{ {g}^{2} + {f}^{2} - c }$