Question

find the centre and radius of circle given
with equation 3x^2+3y^-5x-6y+4 ​

Answer 1

Step-by-step explanation:

We have,

The given equation of circle is $3{x}^{2}+3{y}^{2}-5x-6y+4=0$

$\implies \: {x}^{2}+{y}^{2}- \frac{5}{3} x- \frac{6}{3} y+ \frac{4}{3} =0$

$\implies \: {x}^{2}+{y}^{2}- \frac{5}{3} x- 2y+ \frac{4}{3} =0$

Here, $g=-\frac{5}{6}$ and $f=-1$

So,

center be C

$C \equiv( - g ,- f)$

$C \equiv \bigg( \frac{5}{6} ,1\bigg)$

Radius be R,

$R = \sqrt{ \bigg( \frac{5}{6} \bigg)^{2} + (1)^{2} - \frac{4}{3} }$

$\implies \: R = \sqrt{ \frac{25}{36} + 1 - \frac{4}{3} }$

$\implies \: R = \sqrt{ \frac{25}{36} + \frac{3 - 4}{3} }$

$\implies \: R = \sqrt{ \frac{25}{36} - \frac{1}{3} }$

$\implies \: R = \sqrt{ \frac{25 - 12}{36} }$

$\implies \: R = \sqrt{ \frac{13}{36} }$

$\implies \: R = \frac{ \sqrt{13} }{6}$

Answer 2

Centre = (5/6, 1) and Radius = √(13) / 6.

Step-by-step explanation:

Given :

Equation of circle ;

$3 {x}^{2} + 3 {y}^{2} - 5x - 6y + 4 = 0$

To find :

The centre and radius of the circle.

Solution :

Divide throughout by 3 we have,

${x}^{2} + {y}^{2} - \frac{5}{3}x - 2y + \frac{4}{3} = 0$

comparing it with,

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

Here,

$g = - \frac{5}{6} \: \: and \: \: f = - 1 \: \: \: and \: \: \: c \: \: = \frac{4}{3}$

Centre = (-g, -f)

i.e.

x co-ordinate of the centre = - (coefficient of x)/2

y co-ordinate of the centre = - (coefficient of y)/2

Therefore, centre =

$( \frac{5}{6} , 1)$

Radius =

$\sqrt{ {g}^{2} - {f}^{2} - c}$

=

$\sqrt{ {( \frac{ - 5}{6} )}^{2} + {( - 1)}^{2} - \frac{4}{3} } \\ = \sqrt{ \frac{25}{36} + 1 - \frac{4}{3} } \\ = \sqrt{ \frac{25 + 36 - 48}{36} } \\ = \frac{1}{6} \sqrt{13}$

Hence, centre = (5/6, 1) and radius = √(13) / 6.