The general form of the equation of a circle is x2+y2−4x−8y−5=0. What are the coordinates of the center of the circle?

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1 Answer
Y
May 11, 2017

The center of the circle is at #(2,4)# and the circle has a radius of #5#.

Explanation:

Formatted question: #x^2+y^2-4x-8y-5=0#

Note: The standard form a circle is #(x-h)^2+(y-k)^2=r^2#, where #(h,k)# is the center of the circle and #r# is the radius of the circle.

To find the standard form of the circle, we need to complete the square and move the constants to the right side:
#x^2+y^2-4x-8y=5#
#(x-2)^2-4+(y-4)^2-16=5#
#(x-2)^2+(y-4)^2=25#

Therefore, the center of the circle is at #(2,4)# and the circle has a radius of #5#.

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