Find the center and radius of the circle whose equation is x^2+3x+y^2-17 = 0. The center of the circle is ( , ). The radius of the circle is

$$\begin{array}{rll}
x^2+3x+y^2-17 &=& 0\\\\
x^2+3x+y^2 &=& 17\\\\
x^2+3x+(\frac{3}{2})^2\;\;+y^2 &=& 17+(\frac{3}{2})^2\\\\
(x+\frac{3}{2})^2\;\;+y^2 &=& 17+\frac{9}{4}\\\\
(x+\frac{3}{2})^2\;\;+y^2 &=& \frac{77}{4}\\\\
(x+\frac{3}{2})^2\;\;+y^2 &=& \left(\frac{\sqrt{77}}{2}\right)^2\\\\
\end{array}\\\\
centre\;\;(\frac{-3}{2},\;0)\qquad radius&=& \frac{\sqrt{77}}{2}$$

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Get this in the form (x - xc)² + (y - yc)² = r²

 

First look at the x-terms

x² + 3x can be written as (x + 3/2)² - 9/4  = (x - -3/2)² - 9/4  giving xc = -3/2

 

Now look at the y-terms

y² can be written as (y - 0)² giving yc = 0

 

(xc, yc) = (-3/2, 0)

 

So the original equation can be written as (x + 3/2)² - 9/4 + (y - 0)² - 17 = 0

or (x + 3/2)² + (y - 0)² = 17 +9/4

or (x + 3/2)² + (y - 0)² = 77/4

 

so r² = 77/4

r = (√77)/2

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