The equation of a circle is #x²+y²+2x-4y-8=0#, what are the co-ordinates of the center and the length of the radius of the circle?

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Answer 1 · 2016-05-05T17:31:52

Let's first convert to standard form by doing a double completion of square. This form is much easier to work with.

#x^2 + 2x + y^2 - 4y = 8#

#1(x^2 + 2x + 1 - 1) + (y^2 - 4y + 4 - 4) = 8#

#(x + 1)^2 - 1 + (y - 2)^2 - 4 = 8#

#(x + 1)^2 + (y - 2)^2 = 13#

In standard form #a = (x - p)^2 + (y - q)^2#, the center is at #(p, q)# and the radius is at #a#.

Therefore, the center is at (-1, 2) and the radius measures 13 units.

Hopefully this helps!