SOLUTION: what is the equation of this line? question: find the equations of the line described. tangent lines to x�+y�-18x-2y+48=0 at points where x=4 im solving it using x�+y�-18y

Question 1085760: what is the equation of this line?
question: find the equations of the line described.
tangent lines to x�+y�-18x-2y+48=0 at points where x=4

im solving it using x�+y�-18y-2y+48=0
at x=4
my solution (trying)
P(4,y)
(4)�+y�-18(4)-2y+48=0
16+y�-72-2y+48=0
=y�+2y
=y(y+2)=0
y=0 | y+2=0 y=-2
is it right?
P(4,0) Q(4,-2) is it right?
x�+y�-18x-2y+48=0
(x�-18x)+(y�-2y)= -48
is it right? now i dont know whats next.

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Your approach is correct to find the y-coordinates of the intersection points, but there is an error in your arithmetic.
When you substitute x=4 into the equation of the circle, you should get
y^2 - 2y - 8 = 0 and this factors as (y-4)(y+2) = 0
Thus the two intersection points are (4,4) and (4,-2)
We use implicit differentiation to find dy/dx, which is the slope of the tangent line to the circle:
2xdx + 2ydy - 18dx -2dy = 0
-2xdx = (2y-18)dy -> dy/dx = (9-x)/(y-1)
We want to find dy/dx at the points (4,4) and (4,-2)
Substituting the values we get dy/dx @ (4,4) = 5/3 and dy/dx @ (4,-2) = -5/3
To find the equations for the lines, we use the point-slope form:
Line 1: y-4 = (5/3)(x-4) -> y = (5/3)x - 8/3
Line 2: y+2 = (-5/3)(x-4) -> y = (-5/3)x + 14/3
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