Question

Find the equation of the circle whose centre is (3,-1) and which cuts off an intercept of length 6 from the line 2x-5y+18=0?

Answer 1

`x^2+y^2-6x+2y-28=0`

Explanation:

The length of perpendicular from `(3,-1)` to `2x-5y+18=0` is

`|(2xx3-5xx(-1)+18)/sqrt(2^2+5^2)|=29/sqrt29=sqrt29`

Now as circle cuts off an intercept of length `6`, it intercepts at a distance of `3` on each side from the foot of the perpendicular.

Hence radius of circle is `sqrt((sqrt29)^2+3^2)=sqrt38`

and equation of circle is

`(x-3)^2+(y+1)^2=38`

or `x^2+y^2-6x+2y-28=0`

graph{(x^2+y^2-6x+2y-28)(2x-5y+18)((x-3)^2+(y+1)^2-0.03)=0 [-9.75, 10.25, -4.04, 6.38]}