Answer:
![((1)/(2),\text{ 1), 2 units}](https://img.qammunity.org/2023/formulas/mathematics/college/flsuvskeqr48ofky24op5hi2whvn34242o.png)
Explanations:
The given equation is:
![x^2+y^2-x-2y-(11)/(4)=\text{ 0}](https://img.qammunity.org/2023/formulas/mathematics/college/u68wf2qlmod3vffn83qoof8qzv9gmie411.png)
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
![x^2-x+y^2-2y\text{ = }(11)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/uv8yd1cl9a797dme4sw5duc8v32s13rtem.png)
Add the squares of the half of the coefficients of x and y to both sides of the equation
![\begin{gathered} x^2-x+((-1)/(2))^2+y^2-2y+(-1)^2=(11)/(4)+((-1)/(2))^2+(-1)^2_{} \\ x^2-x+((1)/(2))^2+y^2-2y+1^2=(11+1+4)/(4) \\ (x-(1)/(2))^2+(y-1)^2=(16)/(4) \\ (x-(1)/(2))^2+(y-1)^2=\text{ 4} \\ (x-(1)/(2))^2+(y-1)^2=2^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zgci77jiu34e1223onby7fcatmk600i7ar.png)
Compare the resulting equation with (x - a)² + (y - b)² = r²
![\begin{gathered} \text{The center (a, b) = (}(1)/(2),\text{ 1)} \\ \text{The radius, r = 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6f5c6o53c74fhtyfm5r2g8f558a7smx24w.png)