The equation of a circle is written as x² + 10 + y² - 12y + 1 = 0. What is the location of the center for this circle?

Question

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Answer 1

Need to completed the square for both x and y.

x^2 + 10 + y^2 - 12y +1 = 0

Put all constants on right side.

x^2 + y^2 - 12y = -11

there is really no work with x: x^2 = (x-0)^2 <--- this is the format for the eqtn of a circle

to complete the square with y^2 - 12y take b/2 or -12/2 = -6 this will become (y-6)^2

(y-6)^2 = y^2 -12y + 36 so the left sides's y^2 -12y needs a 36 added to it to make it a perfect square

And whatever you add to the left side must also be added to the right side.

Now we have: (x-0)^2 + y^2 - 12y + 36 = -11 + 36

Putting it as a squares: (x-0)^2 + (y-6)^2 = 25 center of circle is (0,6) radius is √25 = 5