Question

determine the point of intersection of the circle x^2+y^2+6x-4y+8=0 with each of the lines x+1=0. ​

Answer 1

The coordinates of the points of intersection of the line and circle satisfy their equations.

So where the two intersect x= 2y-2 can be substituted in the equation of the circle.

We then get

(2y-2)^2 +y^2 -6(2y-2) +6y+8 = 0

Simplifying we get the quadratic 5y^2 -14y+24=0 whose discriminant b^2–4ac = 196–480 = -284 a negative value.

Therefore, we get imaginary values for y.

Hence the line and the circle do not intersect.