SOLUTION: Find an equation(s) of the circle(s) tangent to y = 0 at (4,0); tangent to 3x - 4y - 17 = 0

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Question 835834: Find an equation(s) of the circle(s) tangent to y = 0 at (4,0); tangent to 3x - 4y - 17 = 0
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
There are two solutions, a big circle and very small one.  We'll
have to do them separately because the y-coordinate of the center
of the big circle is positive and is the same as the radius,
whereas the y-coordinate of the center of the small circle is
negative and therefore the negative of its radius.



Since the circle is tangent to the x-axis at (4,0), we know
that the center (h,k) is on the red vertical line x=4, and
so h=4.

For the big circle, the radius r is the same as k. So the 
center is (4,r). We also know that the radius is the 
perpendicular distance from the center to the given line.

The perpendicular distance from the point (x1,y1)
to the line Ax+y+C=0 is

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

The given line is

3x-4y-17=0, so A=3, B=-4, C=-17, and the distance d
is the radius r, and the point (x1,y1) = (4,r),
the center of the circle.

r = 

r = abs%2812-4r-17%29%2Fsqrt%289%2B16%29

r = abs%28-5-4r%29%2Fsqrt%2825%29

r = abs%28-5-4r%29%2F5

5r = |-5-4r|

Square both sides:

(5r)² = (|-5-4r|)²

25r² = 25+40r+16r²

9r²-40r-25 = 0

(9r+5)(r-5) = 0

r = -5%2F9;  r=5

We ignore the negative solution.

So the equation of the big circle is

(x-h)² + (y-k)² = r²

(x-4)² + (y-5)² = 5²

(x-4)² + (y-5)² = 25.

---------------------------

For the little circle, since it's so tiny, we 
need to zoom in on it:



For the little circle, k, the y-coordinate of the center, is
below the x-axis and is negative.  Therefore k = -r, whereas 
in the big circle k and r were equal.  So the radius r of the 
the little circle is -k. So k = -r, and the center is (4,-r). 
As before we know that the radius is the perpendicular 
distance from the center (4,-r)to the given line.

The perpendicular distance from the point (x1,y1)
to the line Ax+y+C=0 is

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

The given line is

3x-4y-17=0, so as before A=3, B=-4, C=-17, and the distance d
is the radius r, but this time the point (x1,y1) = (4,-r),
the center of the little circle.

r = 

r = abs%2812%2B4r-17%29%2Fsqrt%289%2B16%29

r = abs%28-5%2B4r%29%2Fsqrt%2825%29

r = abs%28-5%2B4r%29%2F5

5r = |-5+4r|

Square both sides:

(5r)² = (|-5+4r|)²

25r² = 25-40r+16r²

9r²+40r-25 = 0

(9r-5)(r+5) = 0

r = 5%2F9;  r = -5

We ignore the negative answer.

So the equation of the little circle is

(x-h)² + (y-k)² = r²

(x-4)² + (y-5%2F9)² = %285%2F9%29%5E2

(x-4)² + (y-5%2F9)² = 25%2F81.

Edwin