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.

*

*From L :4x + 3y + 28 = 0, get $\alpha (= 53.13^0$, but not really necessary). Note further that the blue triangle is a 3-4-5 right triangle.

*Find G where L cuts the y-axis. OG is then known.

*L’ is normal to L and passes through G. The angle between L’ and the y-axis = $\beta = \alpha$.



*A point on the red line will be the center of the required circle. It will cut the y-axis at T, whose co-ordinates are to be determined.

*The blue and green triangles are similar. Therefore, TG is known (using the 3-4-5 properties).

*Find OT and therefore $T = … = (0, \dfrac {-8}{3})$.

*Form the equation of Q which passes through T and has the same slope as L.

*Solve P and Q to get X which is the center of the required circle.

*...
