Answer:
center (0, 3) and radius 2√2
Step-by-step explanation:
Equation of a circle
[tex](x-a)^2+(y-b)^2=r^2[/tex]
where:
- (a, b) is the center
- r is the radius
Given equation:
[tex]x^2+y^2-6y+1=0[/tex]
Subtract 1 from both sides:
[tex]\implies x^2+y^2-6y=-1[/tex]
To create a trinomial with variable y, add the square of half the coefficient of the y term to both sides:
[tex]\implies x^2+y^2-6y+\left(\dfrac{-6}{2}\right)^2=-1+\left(\dfrac{-6}{2}\right)^2[/tex]
[tex]\implies x^2+y^2-6y+9=8[/tex]
Factor the trinomial with variable y:
[tex]\implies x^2+(y^2-6y+9)=8[/tex]
[tex]\implies x^2+(y-3)^2=8[/tex]
Factor [tex]x^2[/tex] to match the general form for the equation of a circle:
[tex]\implies (x-0)^2+(y-3)^2=8[/tex]
Compare with the general form of the equation for a circle:
[tex]\implies a=0[/tex]
[tex]\implies b=3[/tex]
[tex]\implies r^2=8 \implies r=2\sqrt{2}{[/tex]
Therefore, the center is (0, 3) and the radius is 2√2
