Question 258001: y-9=0
x^2+y^2-24x-10y+153=0
Answer by Greenfinch(383)   (Show Source):
You can put this solution on YOUR website! x^2 - 24x + y^2 - 10y + 153 = 0 is a circle with centre at 12,5
So rearranging
(x^2 - 24x + 144) + (y^2 - 10y + 25) + 153 - 169 = 0 completing the square
(x - 12)^2 + (y - 5)^2 = 16 = 4^2 Centre is there at 12, 5 and the radius is 4
Combining with the line y = 9 gives
(x - 12)^2 + (y - 5)^2 - 16 =0 and y = 9
(x - 12)^2 + 4^2 - 16 = 0
(x - 12) ^2 = 0
x = 12 twice, so y = 9 is a tangent touching the circle at x = 12
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