Question

find the radius and centre of the circle x²+y²-4x+6y=12​

Answer 1

Step-by-step explanation:

The given equation of circle is

${x}^{2} + {y}^{2} - 4x + 6y - 12 = 0$

General equation of circle,

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

On comparing,

$g = - 2,f = 3,c = - 12$

We know,

centre$\equiv(-g, -f)$ and radius$= \sqrt{g^{2}+f^{2}-c}$

Now,

$C \equiv(2, - 3)$

And,

$r = \sqrt{4 + 9 + 12}$

$\implies \: r = \sqrt{25} = 5$