Find the center and radius of the circle x2+y2+4x-6y=12

(x + 2)² + (y - 3)² = 5²

Center: (-2, 3)

Radius: 5

Step-by-step explanation:

the form the current equation is in is general form (x² + y² + Dx + Ey + F = 0). 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 + 4x - 6y = 12 <-- the variables are already on one side of the equation, so we do not need to do anything else to this ste[

step two: combine like variables

x² + y² + 4x - 6y = 12 becomes (x² + 4x) + (y² - 6y) = 12

step three: complete the square

we will complete the square of both (x² + 4x) and (y² - 6y)

using the formula (b/2)², with 4 being b in this case we have the following:

(4/2)² = (2)² = 4

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 + 4x + 4) + (y² - 6y + 9) = 12 + 4 + 9 = 25 <-- we add 4 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 + 4x + 4) + (y² - 6y + 9) = 25

(x² + 4x + 4) becomes (x + 2)²

(y² - 6y + 9) becomes (y - 3)²

(x + 2)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 -2, and k is 3 (h is -2 because the original equation has it as x minus h, not plus h)

from this equation, we can see that the center of the circle is (-2 , 3) and to find the radius, we would square 25 to get r²

√25 = 5

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

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step-by-step explanation: