Question

x^2+y^2+4x-6y+12=0 in this equation find the radius and center

Answer 1

Answer:

$center = ( - \frac{1}{2} cofficient \: \: of \: \: x \: , \: - \frac{1}{2} cofficient \: \: of \: \: y) \\ \\ center = ( - 2 \: , \: 3)$

$radius = \sqrt{ {( - 2)}^{2} + {(3)}^{2} - 12 } \\ \\ radius = \sqrt{4 + 9 - 12} = 1$