Question

find the centre and radius of the circle
${3x }^{2} + {3y}^{2} - 5x - 6y + 4 = 0$​

Answer 1

Step-by-step explanation:

General equation of circle is.

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

Given equation.

${3x }^{2} + {3y}^{2} - 5x - 6y + 4 = 0 \\ \\ {x}^{2} + {y}^{2} - \frac{5}{3} x - \frac{6}{3} y + \frac{4}{3} = 0 \\ \\ now \: \: compare \: \: with \: \: general \: eq. \: \: of \: \: circl \\ we \: \: get \\ \\ g = - \frac{5}{</strong><strong>2</strong><strong>×</strong><strong>3}</strong><strong>\: \: \: f = - \frac{6}{</strong><strong>2</strong><strong>×</strong><strong>3} = - </strong><strong>1</strong><strong>$

centere of circle is (-g,-f)

Hence (5/6 , 1 ) is the center.