Question

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

Answer 1

see explanation

Explanation:

The equation of a circle in general form is

x² + y² + 2gx + 2fy + c = 0

with centre = (- g, - f) and radius =
`√(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 =
`(3)/(2)`

2f = - 4 ⇒ f = - 2 and c = 6, thus

centre = ( -
`(3)/(2)`, 2 ) and

radius =
`√((3/2)^2+(-2)^2-6)`

=
`\sqrt{(9)/(4)+4-6 }` =
`\sqrt{(1)/(4) }` =
`(1)/(2)`