Answer:
center = (-2, 6)
radius = 6
Step-by-step explanation:
Equation of a circle
[tex](x-a)^2+(y-b)^2=r^2[/tex]
(where (a, b) is the center and r is the radius)
Therefore, we need to rewrite the given equation into the standard form of an equation of a circle.
Given equation:
[tex]x^2+y^2+4x-12y+4=0[/tex]
Collect like terms and subtract 4 from both sides:
[tex]\implies x^2+4x+y^2-12y=-4[/tex]
Complete the square for both variables by adding the square of half of the coefficient of [tex]x[/tex] and [tex]y[/tex] to both sides:
[tex]\implies x^2+4x+\left(\dfrac{4}{2}\right)^2+y^2-12y+\left(\dfrac{-12}{2}\right)^2=-4+\left(\dfrac{4}{2}\right)^2+\left(\dfrac{-12}{2}\right)^2[/tex]
[tex]\implies x^2+4x+4+y^2-12y+36=-4+4+36[/tex]
Factor both variables:
[tex]\implies (x+2)^2+(y-6)^2=36[/tex]
Therefore:
- center = (-2, 6)
- radius = √36 = 6
