Question

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

Answer 1

The equation of a circle with centre $(h,k)$ and Radius $r$ is given as
$(x-h{)}^2+(y-k{)}^2=r^2$
Here, centre $(h,k)=(\frac{1}{2},\frac{1}{4})$ and radius $\left(r\right)=\frac{1}{12}$
The equation of the circle is
$\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{1}{4})}^2={\left(\frac{1}{12}\right)}^2$
$\Rightarrow x^2-2.x.\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+y^2-2.y.\frac{1}{4}+{\left(\frac{1}{4}\right)}^2=\frac{1}{144}$
$\Rightarrow x^2-x+\frac{1}{4}+y^2-\frac{1}{2}y+\frac{1}{16}=\frac{1}{144}$
$\Rightarrow x^2+y^2-x-\frac{1}{2}y+\frac{1}{4}+\frac{1}{16}-\frac{1}{144}=0$
$\Rightarrow \frac{144x^2+144y^2-144x-72y+36+9-1}{144}=0$
$\Rightarrow {144}^2+144y^2-144x-72y+44=0$
$\Rightarrow 4(36x^2+36y^2-36x-18y+11)=0$
$\Rightarrow 36x^2+36y^2-36x-18y+11=0$
$\therefore$ Therefore, the equation of the required circle is $36x^2+36y^2-36x-18y+11=0$