Answer:[tex](\frac{1}{2},\text{ 1), 2 units}[/tex]Explanations:
The given equation is:
[tex]x^2+y^2-x-2y-\frac{11}{4}=\text{ 0}[/tex]
The standard equation of a circle is given as:
(x - a)² + (y - b)² = r²
where (a, b) is the center
r is the radius
Express the given equation in form of the standard equation
Collect like terms
[tex]x^2-x+y^2-2y\text{ = }\frac{11}{4}[/tex]
Add the squares of the half of the coefficients of x and y to both sides of the equation
[tex]\begin{gathered} x^2-x+(\frac{-1}{2})^2+y^2-2y+(-1)^2=\frac{11}{4}+(\frac{-1}{2})^2+(-1)^2_{} \\ x^2-x+(\frac{1}{2})^2+y^2-2y+1^2=\frac{11+1+4}{4} \\ (x-\frac{1}{2})^2+(y-1)^2=\frac{16}{4} \\ (x-\frac{1}{2})^2+(y-1)^2=\text{ 4} \\ (x-\frac{1}{2})^2+(y-1)^2=2^2 \end{gathered}[/tex]
Compare the resulting equation with (x - a)² + (y - b)² = r²
[tex]\begin{gathered} \text{The center (a, b) = (}\frac{1}{2},\text{ 1)} \\ \text{The radius, r = 2} \end{gathered}[/tex]
