Question

2x^2+2y^2+6x-8y+12=0

Answer 1

Answer:

see explanation

Step-by-step explanation:

The equation of a circle in general form is

x² + y² + 2gx + 2fy + c = 0

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

Given

2x² + 2y² + 6x - 8y + 12 = 0 ← divide through by 2

x² + y² + 3x - 4y + 6 = 0 ← in general form

Compare like terms

2g = 3 ⇒ g = $\frac{3}{2}$

2f = - 4 ⇒ f = - 2 and c = 6, thus

centre = ( - $\frac{3}{2}$, 2 ) and

radius = $\sqrt{(3/2)^2+(-2)^2-6}$

          = $\sqrt{\frac{9}{4}+4-6 }$ = $\sqrt{\frac{1}{4} }$ = $\frac{1}{2}$