Acircle has the equation x2 + y2 – x + 10y + 13 = 0. what is the radius of this circle?

Hint:  Make your "x2 + y2 – x + 10y + 13 = 0" clearer by using " ^ " to indicate exponentiation:

x^2 + y^2 – x + 10y + 13 = 0

Rearrange this so that x terms are together and y terms are also (separately):

x^2  - 1x + y^2 + 10y + 13 = 0

Complete the square twice:  once for x and once for y:

(x^2 - 1x + (-1/2)^2 - (-1/2)^2 + y^2 + 10y + 25 - 25 = - 13

  Condense:

    (x-1/2)^2 + (y+5)^2  - 1/4 - 25 = -13

Combine -1/4, -25 and -13:

   (x-1/2)^2 + (y+5)^2  - 1/4 - 25 = -13 + 1/4 + 25  =  11 3/4 = 47/4

Then we have    (x-1/2)^2 + (y+5)^2  =  47/4, and r^2 = 47/4.  Then the radius of the circle is sqrt(47/4) = sqrt(47)/2  (answer)

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