Question

The centre of a circle is $(-2,3)$ and it touches the straight line $4x+3y+2=0$. Then find the equation of the circle.

Answer 1

Given the centre of the circle is $(-2,3)$.

Also given the circle touches the line $4x+3y+2=0$......(1).

So line (1) is a tangent to the circle.

So the distance between the centre $(-2,3)$ and the line (1) is the radius of the circle.

So the radius is $\left|\frac{4.-2+3.3+2}{\sqrt{4^2+3^2}}\right|=\frac{3}{5}$ units.

Now, the equation of the circle is

$(x+2{)}^2+(y-3{)}^2={\left(\frac{3}{5}\right)}^2$.