Question

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

Answer 1

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