Question

If the radius of the circle 3x square +3ysquare-6x+4y-4=0

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Radius=\frac{5}{3}\:unit}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {3x}^{2} + 3 {y}^{2} - 6x + 4y - 4 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Radius =?$

• According to given question :

$\bold{Eqn \: of \: standard \: circle : } \\ \tt: \implies {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 - - - - - (1)\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {3x}^{2} + {3y}^{2} - 6x + 4y - 4 = 0 \\ \\ \text{Dividing \: both \: side \: by \: 3} \\ \tt: \implies \frac{ {3x}^{2} }{3} + \frac{ {3y}^{2} }{3} - \frac{6x}{3} + \frac{4y}{3} - \frac{4}{3} = 0 \\ \\ \tt: \implies {x}^{2} + {y}^{2} - 2x + \frac{4y}{3} - \frac{4}{3} = 0 - - - - - (2)\\ \\ \text{Comparing \: (1) \: and \: (2)} \\ \tt \circ \: g = - 1 \: \: \: \: \: \: f = \frac{2}{3} \: \: \: \: \: \: c = \frac{ - 4}{3} \\ \\ \bold{For \: Radius : } \\ \tt: \implies Radius = \sqrt{ {g}^{2} + {f}^{2} - c} \\ \\ \tt: \implies Radius = \sqrt{ { (- 1)}^{2} + { (\frac{2}{3} )}^{2} + \frac{4}{3} } \\ \\ \tt: \implies Radius = \sqrt{1 + \frac{4}{9} + \frac{4}{3} } \\ \\ \tt: \implies Radius = \sqrt{ \frac{9 + 4 + 12}{9} } \\ \\ \tt: \implies Radius = \sqrt{ \frac{25}{9} } \\ \\ \green{\tt: \implies Radius = \frac{5}{3} \: unit}$

Answer 2

Given :-

$\displaystyle{\sf{3x^2+3y^2-6x+4y-4=0.}}$

To find :-

$\textsf{The radius of the circle.}$

Solution :-

$\textsf{General equation of a circle = }\sf{x^2+y^2+2gx+2fy+c=0.}$

$\textsf{So, the coefficient of }\sf{x^2\ and\ y^2\ must\ be\ equal,\ which\ is\ 3\ here.}$

$\textsf{Coefficient of xy must be 0, where in this equation, it is neglected.}$

$\textsf{Thus, }\sf{3x^2+3y^2-6x+4x-4=0}$

$\underline{\textsf{Dividing the equation by 3:}}$

$\displaystyle{\sf{\implies \frac{3x^2}{3}+\frac{3y^2}{3}-\frac{6x}{3} +\frac{4x}{3}-\frac{4}{3}=0 }}\\\\\displaystyle{\sf{\implies x^2+y^2-2x+\frac{4x}{3}-\frac{4}{3}=0 }}$

$\textsf{Now, when we compare this result with the general equation,}\\\textsf{we get:}$

$\bold{\sf{g=-1,\ f=\frac{2}{3}\ and\ c=\frac{-4}{3}. }}$

$\underline{\bold{The\ radius\ of\ the\ circle\ :}}$

$\sf{Radius=\sqrt{g^2+f^2-c} }\\\\\sf{Radius=\sqrt{(-1)^2+(\frac{2}{3} )^2 -\frac{-4}{3} }$

$\sf{Radius=\sqrt{1+\frac{4}{9}+\frac{4}{3} } }$

$\sf{Radius=\sqrt{\frac{9+4+12}{9} }}\\\\\sf{Radius=\sqrt{\frac{25}{9} }}$

$\underline{\boxed{\sf{Radius=\frac{5}{3}\ units }}}$