SOLUTION: Find an equation(s) of the line(s) containing (5,4) and at a distance 2 from (-1,-3).

Question 672756: Find an equation(s) of the line(s) containing (5,4) and at a distance 2 from (-1,-3).
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find an equation(s) of the line(s) containing (5,4) and at a distance 2 from (-1,-3).
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The lines are tangents to a circle of radius 2 centered at (-1,-3)
The distance from (5,4) to the center (-1,-3) = sqrt(85).
Right angles are formed at the tangent points.
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The distance from (5,4) to the tangent points = 9.
The tangent points are the intersection of the circle above and a circle of radius 9 centered at (5,4).
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----------------------------------- Subtract
12x + 14y = -46
6x + 7y = -23 is an equation of the line thru the 2 tangent points.
y = (-6x - 23)/7
Sub for y in one of the circles

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=63504 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.764705882352941, -2.2. Here's your graph:

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x = -2.2 --> y = (-6*-2.2 - 23)/7 = -1.4
Tangent point at (-2.2,-1.4)
Equation of the line thru (-2.2,-1.4) and (5,4) is
3x - 4y = -1
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x = 0.7647 --> y = -3.94117 --> tangent point at (0.7647,-3.84117)
Equation of line thru the 2 points is
7.94x - 4.2353y = 22.7588 (approximation)
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