SOLUTION: The following equation is the general form of the circle
2x^2 - 5x + 2y^2 - 7y = 0.
Complete the square to write the equation of the given circle in standard form to determine
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-> SOLUTION: The following equation is the general form of the circle
2x^2 - 5x + 2y^2 - 7y = 0.
Complete the square to write the equation of the given circle in standard form to determine
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Question 896287: The following equation is the general form of the circle
2x^2 - 5x + 2y^2 - 7y = 0.
Complete the square to write the equation of the given circle in standard form to determine the center (h,k) and the radius r.
I figured out that the standard equation is (x-1.25)^2+(y-1.75)^2 = 0, but what is the radius? Answer by Theo(13342) (Show Source):
it's probably easiest to remove the coefficients of the x^2 term and the y^2 term in one shot because the coefficients are the same.
divide both sides of this equation by 2 to get:
x^2 - 5/2x + y^2 - 7/2y = 0
now break the equation up into 2 parts with one part containing the x terms and the other part containing the y terms.
you get:
(x^2 - 5/2x) + (y^2 - 7/2y) = 0
now complete the square on each term.
you get:
(x - 5/4)^2 - (5/4)^2 + (y - 7/4)^2 - (7/4)^2 = 0
now add (5/4)^2 and (7/4)^2 to both sides of the equation to get:
(x-5/4)^2 + (y-7/4)^2 = (5/4)^2 + (7/4)^2
that's the part you were missing !!!!!.
simplify to get:
(x-5/4)^2 + (y-7/4)^2 = 37/8
the center of your circle should be (5/4,7/4) and the radius of your circle should be sqrt(37/8) if we did this correctly.
let's see if we did.
here's a graph of your original equation.
the points on the circle starting from top and rotating in a clockwise direction are:
1.25, 3.901
3.401, 1.75
1.25, -.401
-.901, 1.75
all the points on the circle are calculated by adding sqrt(37/8) to get from the center of the circle to the surface of the circle.
the calculations all check out ok.
the radius of the circle is sqrt(37/8).
it appears your only error was in not adding the byproduct of the completing the square process to both sides of the equation which effectively adds them to the right side of the equation.
