Equations of a line tangent to a circle

[Subpar]

1. The problem statement, all variables and given/known data
Given the circle (x+1)^2 + (y-3)^2 = 25, determine the equations of the tangents to the circle with the slope -3/4.


2. Relevant equations
y = mx + b


3. The attempt at a solution
I thought that if I could find the equation of the line that passed through the center of the circle and had a slope perpendicular to -3/4 (4/3) I could then use the equation to find points on the circle which a tangent with a slope -3/4 touched and solve from there. However once I began doing this I started getting a bizarre number and stopped. I have no problem forming the equation of a tangent when given a point on the circle, but I can't figure out how to solve the question when only given the slope. Any help would be appreciated.

The textbook gives the answers as: 3x + 4y = 34, 3x + 4y = -16

[Mentallic]

Show us what you've done and we'll be able to point out where your error lies, because since you seem to know the method in solving the problem, the issue probably lies in your algebra.

[mege]

To solve geometrically, draw out the circle and your perpindicular line. The line intersecting the circle should form a 3,4,5 triangle (remember, radius 5 = hypotenuse). You can use this information to find the points.

I can't recall of an easy way to solve this algebraically.

[eshaw]

You were on the right track. Once you find the line with slope 4/3 that goes through the center of the circle at (-1, 3), you need to plug that equation back into equation for the circle and use the quadratic formula to find the two values of x. Then plug the values of x that you got into the equation for the line that goes through the center to find the two values of y. You should get two coordinates through which the tangent lines go through, which are (2, 7) and (-4, -1). From there it is easy to find that 3x+4y=34 and 3x+4y=-16 are the two lines you're looking for.