Question

Find the equation to the circle which has its centre at the point $(3,4)$ and touches the straight line $5x+12y=1$.

Answer 1

Perpendicular distance of the centre from the line will be equal to the radius as the line is tangent to the circle

$\therefore d=\frac{5\left(3\right)+12\left(4\right)-1}{13}=r$

$d=r=\frac{62}{13}$

So, the equation of circle will be

$(x-3{)}^2+(y-4{)}^2={\left(\frac{62}{13}\right)}^2$

$x^2+y^2-6x-8y+\frac{381}{169}=0$