SOLUTION: Find an equation of a circle tangent to 3x+4y-16=0 at (4,1) with a radius of 5.

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Question 783735: Find an equation of a circle tangent to 3x+4y-16=0 at (4,1) with a radius of 5.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find an equation of a circle tangent to 3x+4y-16=0 at (4,1) with a radius of 5.
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The line thru the center of the circle is perpendicular to the given line, and the center is 5 units from the tangent points.
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Find the slope of the line
3x+ 4y = 16
solve for y
y = -3x/4 + 4
Slope of the line = -3/4
Slope of the perpendicular line = 4/3
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Find the eqn of the perpendicular line thru (4,1)
use y = mx + b and the point to find b, the y-intercept
1 = (4/3)*4 + b
b = -13/3
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The perpendicular line is y = 4x/3 - 13/3
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There are 2 points 5 units from (4,1) on y = 4x/3 - 13/3 --> 2 tangent circles
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Draw a circle radius = 5 centered at (4,1)

y = 4x/3 - 13/3
Sub for y

Solve for x



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=22500 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 7, 1. Here's your graph:

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x = 7, y = 5 --> (7,5)
--> is one circle
========================================
x = 1, y = -3 --> (1,-3)
is the 2nd circle

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