Question

a circle is defined by the equation given below.
$x^{2}+y^{2}-x - 2y-\frac{11}{4}=0$
what are the coordinates for the center of the circle and the length of the radius?
a. $(-\frac{1}{2},-1)$, 4 units
b. $(\frac{1}{2},1)$, 2 units
c. $(-\frac{1}{2},-1)$, 2 units
d. $(\frac{1}{2},1)$, 4 units

Explanation:

Step1: Convert to standard form

The general equation of a circle is \((x - a)^2+(y - b)^2=r^2\), where \((a,b)\) is the center and \(r\) is the radius. Given \(x^{2}+y^{2}-x - 2y-\frac{11}{4}=0\). Complete the square for \(x\) and \(y\) terms.
For \(x\): \(x^{2}-x=(x-\frac{1}{2})^{2}-\frac{1}{4}\)
For \(y\): \(y^{2}-2y=(y - 1)^{2}-1\)
The equation becomes \((x-\frac{1}{2})^{2}-\frac{1}{4}+(y - 1)^{2}-1-\frac{11}{4}=0\)

Step2: Simplify the equation

\((x-\frac{1}{2})^{2}+(y - 1)^{2}-\frac{1 + 4+11}{4}=0\)
\((x-\frac{1}{2})^{2}+(y - 1)^{2}=4\)

Answer:

B. \((\frac{1}{2},1)\), \(2\) units