Answer:
Coordinates of center is [tex](\frac{1}{2},1)[/tex] and length of the radius is 2 units
Step-by-step explanation:
Given Equation of circle : [tex]x^2+y^2-x-2y-\frac{11}{4}=0[/tex]
We have to find coordinates of center and length of the radius.
consider,
[tex]x^2+y^2-x-2y-\frac{11}{4}=0[/tex]
[tex]x^2-x+y^2-2y-\frac{11}{4}=0[/tex]
using completing the square method, we get
[tex]x^2-x+\frac{1}{4}-\frac{1}{4}+y^2-2y+1-1-\frac{11}{4}=0[/tex]
[tex](x-\frac{1}{2})^2-\frac{1}{4}+(y-1)^2-1-\frac{11}{4}=0[/tex]
[tex](x-\frac{1}{2})^2+(y-1)^2-\frac{1}{4}-1-\frac{11}{4}=0[/tex]
[tex](x-\frac{1}{2})^2+(y-1)^2+\frac{-1-4-11}{4}=0[/tex]
[tex](x-\frac{1}{2})^2+(y-1)^2-\frac{16}{4}=0[/tex]
[tex](x-\frac{1}{2})^2+(y-1)^2=\frac{16}{4}[/tex]
[tex](x-\frac{1}{2})^2+(y-1)^2=2^2[/tex]
Now comparing with the standard form of the circle, (x-h)² + (y-k)² = r²
we get [tex](h,k)=(\frac{1}{2},1)[/tex] and r = 2
Therefore, Coordinates of center is [tex](\frac{1}{2},1)[/tex] and length of the radius is 2 units
