Radius-4 circle skewered on one line and touching another Find the equation of the circle with radius $4$ units, whose centre lies on the line $4x+13y=32$ and which touches the line $4x+3y+28=0$.
My Approach:
Radius $r=4$ units
Let $P(h,k)$ be the centre of the circle. Then $4h+13k=32$.
Please help me to move further.

$4x+13y=32\iff \dfrac y4=\dfrac{8-x}{13}=m$(say)
So, any point on $4x+13y=32$ will be $(8-13m,4m)$
Now the perpendicular distance of $(8-13m,4m)$ from $4x+3y+28=0$ will be $$\dfrac{|4(8-13m)+3(4m)+28|}{\sqrt{4^2+3^2}}$$ which needs to be $=$ radius i.e., $4$ unit
