SOLUTION: For what value of k , line 3x+4y+k=0 is tangent to the circle x^2+y^2+6x-y-1=0? a)13 a)19 c)25 d)30

Question 1083556: For what value of k , line 3x+4y+k=0 is tangent to the circle x^2+y^2+6x-y-1=0?
a)13
a)19
c)25
d)30
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
For what value of k , line 3x+4y+k=0 is tangent to the circle x^2+y^2+6x-y-1=0?
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3x+4y+k=0
y = (-3/4)x - k/4
Slope m = -3/4
================
x^2+y^2+6x-y-1=0
x^2 + 6x + 9 + y^2 - y + 1/4 = 1 + 9 + 1/4 = 10.25

The center is (-3,1/2)
Find the line thru the center perpendicular to the given line.
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slope = -1/m = 4/3
y- 1/2 = (4/3)*(x+3)
The 2 tangent points are the intersection of y- 1/2 = (4/3)*(x+3) and the circle.
----
2y- 1 = (8/3)*(x+3) = 8x/3 + 8
y = 4x/3 + 9/2
---
x^2+y^2+6x-y-1=0
Sub for y
x^2 + (4x/3 + 9/2)^2 + 6x - (4x/3 + 9/2) - 1 = 0
x^2 + 16x^2/9 + 12x + 81/4 + 6x - 4x/3 - 9/2 - 1 = 0
25x^2/9 + 50x/3 + 59/4 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=113.88888888889 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -1.07906272877014, -4.92093727122987. Here's your graph:

==================
find the y values, then the equations of the lines thru the 2 points with a slope of -3/4.
then find 2 values for k.

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