How do you find the center and radius for x^2+y^2 +2x-4y=4x2+y2+2x−4y=4?

1 Answer
sente
May 30, 2016

Put the equation into standard form to find it to be a circle with radius 33 and center (-1, 2)(−1,2)

Explanation:

In order to put the equation of a conic section into standard form (at which point we can see most of its properties), we use a process called completing the square on both variables.

x^2+y^2+2x-4y=4x2+y2+2x−4y=4

=>(x^2+2x+1) - 1 + (y^2 - 4y + 4) - 4 = 4⇒(x2+2x+1)−1+(y2−4y+4)−4=4

=> (x+1)^2 + (y-2)^2 = 4+4+1 = 9⇒(x+1)2+(y−2)2=4+4+1=9

The standard form of the equation of a circle with radius rr centered at (x_0, y_0)(x0,y0) is

(x-x_0)^2 + (y-y_0)^2 = r^2(x−x0)2+(y−y0)2=r2

Thus, as our equation may be written as

(x-(-1))^2+(y-2)^2=3^2(x−(−1))2+(y−2)2=32

it represents a circle of radius 33, centered at (-1,2)(−1,2)

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