Question

Find the centre and the radius of the
circle 3x² + 3y²-5x-6y+4 = 0​

Answer 1

Given equation is

$\sf3x² + 3y²-5x-6y+4 = 0$

$\small\fbox\red{Divide throughout by 3 :}$

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

Now comparing above equation with general equation of circle,

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

$\small\fbox\green{Centre of the circle is (-g,-f)=(5/6,1) and radius}[\tex]</p><p></p><p>[tex] \sf\sqrt{g {}^{2} + f {}^{2} - c } \implies \sf\sqrt{ \frac{25}{36} + 1 - \frac{4}{3} }= \sqrt{ \frac{13}{6} }$