Question 672756
Find an equation(s) of the line(s) containing (5,4) and at a distance 2 from (-1,-3).
-----------
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.
----
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).
----------
{{{(x+1)^2 + (y+3)^2 = 4}}}
{{{(x-5)^2 + (y-4)^2 = 81}}}
---
{{{x^2 + y^2 + 2x + 6y = -6}}}
{{{x^2 + y^2 - 10x - 8y = 40}}}
----------------------------------- 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
{{{x^2 + y^2 + 2x + 6y = -6}}}
{{{x^2 + (36x^2 + 276x + 529)/49 + 2x + (-36x - 138)/7 = -6}}}
{{{49x^2 + 36x^2 + 276x + 529 + 98x - 252x - 966 = -294}}}
{{{85x^2 + 122x - 143 = 0}}}
*[invoke solve_quadratic_equation 85,122,-143]
==========================
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
==============================================
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)