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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

User Kalaschni
by
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1 Answer

12 votes

Answer:

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

User Michelangelo
by
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