# given the equation x^2+y^2-18x-12y+113=0 A: write the equation in standard form B:  State the center, radius C:  graph

beckden | High School Teacher | (Level 1) Educator

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This is a circle.  We write it in standard form by completing the square.

Completing the square involves adding half the linear term squared, we need to do that for both the x and y terms.  You could do this either by adding these numbers to both sides or adding and subtracting this number.  When we complete the square (x^2 + bx + (b/2)^2) = (x+b/2)^2

x^2-18x + (-18/2)^2 - (-18/2)^2 + y^2 - 12y + (-12/2)^2 - (-12/2)^2 = -113

or

x^2 - 18x + (-18/2)^2 + y^2 - 12y + (-12/2)^2 = -113 + (-18/2)^2 + (-12/2)^2

We then get

(x - 18/2)^2 + (y-12/2)^2 = -113 + 81 + 36

or finally

(x - 9)^2 + (y - 6)^2 = -113 + 117 = 4

So answer for a) is (x-9)^2 + (y-6)^2 = 4

So knowing the general equation for a circle of radius r and center at (a,b) is (x-a)^2 + (y-b)^2 = r^2

We get the answer for b) Center (9,6) and radius = 2

For c)

Hope that helps....