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