# 3.  Find the equation of the circle with   centre $\left(\frac{1}{2},\frac{1}{4} \right )$ and radius $\frac{1}{12}$

As we know,

The equation of the circle with center ( h, k) and radius r is give by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=\left ( \frac{1}{2},\frac{1}{4} \right )$

AND

$r=\frac{1}{12}$

So the equation of circle is:

$\left ( x-\frac{1}{2}\right )^2+\left ( y-\frac{1}{4}\right )^2=\left ( \frac{1}{12}\right )^2$

$x^2-x+\frac{1}{4}+y^2-\frac{1}{2}y+\frac{1}{16}=\frac{1}{144}$

$x^2+y^2-x-\frac{1}{2}y-\frac{11}{36}=0$

$36x^2+36y^2-36x-18y-11=0$

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