Finding coordinates of the point where the line meets the...

...circle?

Find the coordinates of the point where the circle x² - 8x + y²- 2y - 3 = 0 meets the line x - 2y - 12 = 0

The point where the line meets the circle will satisfy the equation of both the line as well as the circle.

From the equation x - 2y - 12 = 0, we have
x = 2y + 12

Substituting this value of x in the equation of the circle, we have
(2y + 12)^2 - 8(2y + 12) + y^2 - 2y - 3 = 0
=> 4y^2 + 48y + 144 - 16y - 96 + y^2 - 2y - 3 = 0
=> 5y^2 + 30y + 45 = 0
=> y^2 + 6y + 9 = 0
=> (y + 3)^2 = 0
=> y + 3 = 0
=> y = -3

Solving for x, we have x = 2y + 12 = 2(-3) + 12 = 6

So, the point where the line meets the circle is (x, y) = (6, -3)

x - 2y - 12 = 0
2y=x-12
y=x/2-6

Substitute y=x/2 - 6 into the equation of the circle.

x^2-8x+y^2-2y-3=0

x^2-8x+(x/2-6)^2-2(x/2-6) - 3 = 0
x^2 - 8x +x^2/4 -6x+36 -x+12 -3 = 0
4x^2-32x+x^2-24x+144-4x+48-12=0
5x^2-60x+180 = 0
x^2-12x+36=0
(x-6)^2=0
x=6
x-2y-12=0
6-2y-12=0
-2y=6
y=-1

The coordinates of the point where the circle x² - 8x + y²- 2y - 3 = 0 meets the line x - 2y - 12 = 0 is:
(6,-1)