The equation of a circle is x2 + y2 - 12x + 6y + 20 = 0.
what is the radius of the circle? enter your answer in the box. r = units

Center: (6, -3) Radius: 5

Step-by-step explanation:

the form the current equation is in is general form. we need to get the equation into standard form which is (x - h)² + (y - k)² = r², where h and k are the center of the circle, and r² is the radius of the circle

step one: seperate the variables on one side of the equation

x² + y²- 12x + 6y + 20 = 0 <-- subtract 20 from both sides of the equation to get -20 on both sides

x² + y²- 12x + 6y = -20

step two: combine like variables

x² + y²- 12x + 6y = -20 becomes (x² - 12x) + (y² + 6y) = -20

step three: complete the square

we will complete the square of both (x² - 12x) and (y² + 6y)

using the formula (b/2)², with -12 being b in this case we have the following:

(-12/2)² = (-6)² = 36

using the same formula but instead applied to (y² + 6y), with b being 6, we have the following:

(6/2)² = (3)² = 9

the equation should now look like the following:

(x² - 12x + 36) + (y² + 6y + 9) = -20 + 36 + 9 = 25 <-- we add 36 and 9 on both sides because what we do to one side, we do to the other.

step four: factor the equation

we are going to factor the equation below in order to get it into standard form

(x² - 12x + 36) + (y² + 6y + 9) = 25

(x² - 12x + 36) becomes (x - 6)²

(y² + 6y + 9) becomes (y + 3)²

(x - 6)² + (y + 3)² = 25

this looks a lot like the standard form of an equation of a circle (x - h)² + (y - k)² = r², where h is 6, and k is -3 (because the original equation has it as y minus k, not plus k)

from this equation, we can see that the center of the circle is (6 , -3) and to find the radius, we would square 25 to get r²

√25 = 5

to conclude, the final equation is (x - 6)² + (y + 3)² = 5² , with the radius being 5 and the center being (6, -3)

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step-by-step explanation: