find the center and radius for the circle defined by the equation: x2+y2+5x-y+2=0

Question

Find the center and radius for the circle defined by the equation: x2+y2+5x-y+2=0

Answer 1

First, Barbara, we have to group our x's and y's together and move the 2 to the other side

 

x² + 5x + y² - y = -2

 

Now we are going to complete the square twice. First, we take half of 5 (from 5x), square it, and add it to both sides. We do the same thing with -1 (from -y).

 

x² + 5x + 6.25 + y² - y + .25 = -2 + 6.25 + .25

 

The first three terms equal (x + 2.5)² and the last three equal (y - .5)², so, we have...

 

(x + 2.5)² + (y - .5)² = 4.5

 

Remembering that our equation is (x - h)² + (y - k)² = r², our center is (-2.5, .5) and our radius is 2.12, the square root of 4.5.

 

Hope this helps!

Answer 2

x² + 5x + y² - y = -2  Re-arrange the terms. Then complete the square for both x and y.

 

To complete the square, take half of 5 (x's coefficient) and square it. Take half of -1 (y's coefficient) and square it. Then, add those numbers to both sides of the equation.

 

(x² + 5x + 25/4) + (y² - y + 1/4) = -2 + 25/4 + 1/4 = 18/4

 

Write each trinomial as a binomial squared.

(x + 5/2)² + (y- 1/2)² = 9/2

 

Center: (-5/2, 1/2)    Radius = √9/√2 = 3 √2

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