Question 783735
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)
{{{(x-4)^2 + (y-1)^2 = 25}}}
y = 4x/3 - 13/3
Sub for y
{{{(x-4)^2 + (4x/3 - 13/3 -1)^2 = 25}}}
Solve for x
{{{9(x-4)^2 + (4x - 16)^2 = 225}}}
{{{9x^2 - 72x + 144 + 16x^2 - 128x + 256 = 225}}}
{{{25x^2 - 200x + 175 = 0}}}
*[invoke solve_quadratic_equation 25,-200,175]
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x = 7, y = 5 --> (7,5)
--> {{{(x-7)^2 + (y-5)^2 = 25}}} is one circle
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x = 1, y = -3 --> (1,-3)
{{{(x-1)^2 + (y+3)^2 = 25}}} is the 2nd circle