Question

The points of intersection of the line $4x - 3y - 10 = 0 $ and the circle $x^2 + y^2 - 2 x + 4 y - 20 = 0$ are ...and ....

Answer 1

For point of intersection, we put
$x = \frac{3y + 10}{4} $ in $ x^2 + y^2 - 2x +4y - 20 = 0 $
$\Rightarrow \big( \frac{3y + 10}{4}\big)^2 \, + \, y^2 - 2\big(\frac{3y + 10}{4}\big) + 4y - 20 = 0 $
$\Rightarrow 25y^2 + 100y - 300 = 0 $
$\Rightarrow y^2 + 4y - 12 = 0$
$\Rightarrow (y-2) (y+6) = 0$
$\Rightarrow y = -6, 2$
When $ y = -6 \Rightarrow x = -2 $
When $y = 2 \Rightarrow x = 4 $
$ \therefore $ Point of intersection are $(-2, -6) (4,2).$