Question

The equation of a circle is x² + y²-6y+1=0. What are the coordinates of

the center and the length of the radius of this circle?
(1) center (0,3) and radius 2√2
(2) center (0,-3) and radius 2√2
(3) center (0.6) and radius √35
(4) center (0,-6) and radius √35

Answer 1

center (0, 3) and radius 2√2

Explanation:

Equation of a circle

`(x-a)^2+(y-b)^2=r^2`

where:

• (a, b) is the center
• r is the radius

Given equation:

`x^2+y^2-6y+1=0`

Subtract 1 from both sides:

`\implies x^2+y^2-6y=-1`

To create a trinomial with variable y, add the square of half the coefficient of the y term to both sides:

`\implies x^2+y^2-6y+\left((-6)/(2)\right)^2=-1+\left((-6)/(2)\right)^2`

`\implies x^2+y^2-6y+9=8`

Factor the trinomial with variable y:

`\implies x^2+(y^2-6y+9)=8`

`\implies x^2+(y-3)^2=8`

Factor
`x^2` to match the general form for the equation of a circle:

`\implies (x-0)^2+(y-3)^2=8`

Compare with the general form of the equation for a circle:

`\implies a=0`

`\implies b=3`

`\implies r^2=8 \implies r=2√(2){`

Therefore, the center is (0, 3) and the radius is 2√2