Question

Find the points on the straight line ( x-y+1)=0 such that length of tangent from it to the circle $x^2+y^2-3x=0$ is 2.The points are (-1,0) and (p,q) then p+q =

Answer 1

let $P(x_1,y_1)$ be a point on the line $x-y+1=0$ from which the leng the of the taments to the circle $s=x^2+y^2<3x=0$ is of lengthe 2 . Therefore,

$x_1-y_1+1=0\text{and}\sqrt{s_{11}}=2$

$x_1-y_1+1=6\text{and}{x}_1^2+y_1^2-3x_1=4$

Therefore,

Theryore points are $(-1,0),(p,q)=(\frac{3}{2},\frac{5}{2})$

$p+q=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4$