Question

Find the equation of circle which touched the lines $4x-3y+10=0$and $4x-3y-30=0$ and whose centre lies on the line $2x+y=0$

Answer 1

The given tangents are parallel and therefore the distance between them

= 2r = $\frac{10-(-30)}{5}=8\therefore r=4$

Now the diameter 2x + y = 0 cuts the given line in the points P (-1, 2) and Q (3, -6) whose middle point is therefore the centre (1, - 2).

Hence its equation is $(x-1{)}^2+(y+2{)}^2=4^2$