Question

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

Answer 1

Here it is given that centre $(h,k)=(\frac{1}{2},\frac{1}{4})$ and radius $r=\frac{1}{2}$
Therefore the equation of the circle is
${(x-\frac{1}{2})}^2+{(y-\frac{1}{4})}^2={\left(\frac{1}{12}\right)}^2$
$\Rightarrow x^2-x+\frac{1}{4}+y^2-\frac{y}{2}+\frac{1}{16}=\frac{1}{144}$
$\Rightarrow x^2-x+\frac{1}{4}+y^2-\frac{y}{2}+\frac{1}{16}-\frac{1}{144}=0$
$\Rightarrow 144x{}^2-144x+36+144y{}^2-72y+9-1=0$
$\Rightarrow 36x{}^2+36y{}^2-36x-18y+11=0$