SOLUTION: find the equation of a circle tangent to 3x+4y-2=0 with center (4,1)

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Question 938727: find the equation of a circle tangent to 3x+4y-2=0 with center (4,1)
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
We have the center of the circle (h,k) = (4,1), and the
equation of a circle:

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

becomes

%28x-4%29%5E2%2B%28y-1%29%5E2=r%5E2

So all we are lacking is r, the radius.

We sketch the graph approximately:



The green line is a radius, and it is the perpendicular 
distance from the center to the line.  We use this formula:

The perpendicular distance from the point (x1,y1)
to the line Ax+By+C=0 is given by this formula:

d = abs%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29

(x1,y1) = (4,1)

3x+4y-2=0 is Ax+By+C=0, so A=3, B=4, C=-2

r = abs%283%2A4%2B4%2A1-2%29%2Fsqrt%283%5E2%2B4%5E2%29

r = abs%2812%2B4-2%29%2Fsqrt%289%2B16%29

r = abs%2814%29%2Fsqrt%2825%29

r = 14%2F5

So the equation of the circle is:

%28x-4%29%5E2%2B%28y-1%29%5E2=%2814%2F5%29%5E2

Edwin