Question

Find the equation of circle with radius 4 units and whose centre lies on the line `13x+4y=32` and which touches the line `3x+4y+28=0`.?

Answer 1

see explanation.

Explanation:

Let equation of `L1` be `13x+4y=32`, and equation of `L2` be `3x+4y+28=0`.
Let center of the circle be `(h,k)`, as shown in the figure.
given that `(h,k)` lies on `L1`,
Substituting `h=x, y=k` into `13x+4y=32`, we get,
`13h+4k=32`,
`=> k=(32-13h)/4`
recall that the perpendicular distance from a point `P(h,k)` to a line `ax+by+c=0` is `d=|ah+bk+c|/sqrt(a^2+b^2)`,
given that `r=4`, and the circle touches `L2`,
`=> 4=|3h+4xx(32-13h)/4+28|/(sqrt(3^2+4^2))`
`=> 4=|3h+32-13h+28|/5`
`=> 20=|-10h+60|`
`=> h=4 or 8`
when `h=4, k=(32-13xx4)/4=-5`
when `h=8, k=(32-13xx8)/4=-18`
`=> (h,k)=(4,-5), (8,-18)`
equations of the circles are:
`(x-4)^2+(y+5)^2=16`,
and `(x-8)^2+(y+18)^2=16`