\(\frac{\sqrt33}{4}\)
\(\frac{\sqrt5}{6}\)
\(\frac{5}{6}\)
\(\frac{33}{4}\)
Correct answer is A
Standard Form equation of a circle (Center-Radius Form): \((x − a)^2 + (y − b)^2 = r^2\)
Where "a" and "b" are the coordinates of the center and "r" is the radius of the circle
\(2x^2+2y^2-4x+5y+1=0\)
Divide through by 2
= \(x^2+y^2-2x+\frac{5}{2}y+\frac{1}{2}= 0\)
=\(x^2-2x+y^2+\frac{5}{ 2}y=-\frac{1}{ 2}\)
=\(x^2-2x+1^2+y^2+\frac{5}{ 2}y+(\frac{5}{ 4})^2-1-\frac{25}{16}=-\frac{1}{2}\)
=\((x-1)^2+(y+\frac{5}{4})^2=-\frac{1}{2}+1+\frac{25}{16}\)
=\((x-1)^2+(y-(-\frac{5}{4}))^2=\frac{33}{16}\)
=\((x-1)^2+(y-(-\frac{5}{4}))^2=(\frac{\sqrt33}{4})^2\)
\(\therefore a = 1, b = - \frac{5}{4} and\) \(r \frac{\sqrt33}{4} (answer)\)