Asked by Shelby
Put the following equations in standard form. State the center and the radius.
x^2-5x+y^2+4y=-3
I get how to end up with (-2.5,2), what is the radius?
x^2-5x+y^2+4y=-3
I get how to end up with (-2.5,2), what is the radius?
Answers
Answered by
Steve
if you got the center, you already did all the heavy lifting:
(x^2-5x) + (y^2+4y) = -3
(x^2-5x+2.5^2) - 2.5^2 + (y^2+4y+4)-4 = -3
(x-2.5)^2 + (y+2)^2 = -3 + (5/2)^2 + 4
(x-2.5)^2 + (y+2)^2 = 29/4
so, aside from getting the signs wrong for the center, you've done the hard work but forgot to carry the rest of the information during the calculation.
recall the standard equation for a circle of radius r with center at (h,k):
(x-h)^2+(y-k)^2 = r^2
so,the radius is √29/2
(x^2-5x) + (y^2+4y) = -3
(x^2-5x+2.5^2) - 2.5^2 + (y^2+4y+4)-4 = -3
(x-2.5)^2 + (y+2)^2 = -3 + (5/2)^2 + 4
(x-2.5)^2 + (y+2)^2 = 29/4
so, aside from getting the signs wrong for the center, you've done the hard work but forgot to carry the rest of the information during the calculation.
recall the standard equation for a circle of radius r with center at (h,k):
(x-h)^2+(y-k)^2 = r^2
so,the radius is √29/2
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