SOLUTION: Find the center of the circles tangent to 5x-2y-1=0 at 1,2 with a radius of 3

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Question 1141543: Find the center of the circles tangent to 5x-2y-1=0 at 1,2 with a radius of 3
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
If the equation of the circle is
%28x-h%29%5E2%2B%28y-k%29%5E2=9 where (h,k) is the center, we can find the gradient by differentiation:
2%28x-h%29%2B2%28y-k%29%28dy%2Fdx%29=0
At (1,2) the gradient dy%2Fdx is given by
1-h%2B%282-k%29%28dy%2Fdx%29=0
Therefore dy%2Fdx=%28h-1%29%2F%282-k%29

Point (1,2) lies on the circle:
%281-h%29%5E2%2B%282-k%29%5E2=9
The gradient of the line 5x-2y-1=0 is given by
dy%2Fdx: 5-2%28dy%2Fdx%29=0

dy%2Fdx=5%2F2

This gradient is tangent to the circle, so
%28h-1%29%2F%282-k%29=5%2F2
2h-2=10-5k
+2h=12-5k
h=6-5k%2F2


So,
%281-6%2B5k%2F2%29%5E2%2B%282-k%29%5E2=9
25%28k%2F2-1%29%5E2%2B4-4k%2Bk%5E2-9=0
25%28k%5E2%2F4-k%2B1%29-5-4k%2Bk%5E2=0
25k%5E2%2F4-25k%2B25-5-4k%2Bk%5E2=0
29k%5E2%2F4-29k%2B20=0
%2829%2F4%29%28k%5E2-4k%29=-20
+%2829%2F4%29%28k%5E2-4k%2B4-4%29=-20
%2829%2F4%29%28%28k-2%29%5E2-4%29=-20
%2829%2F4%29%28k-2%29%5E2-29=-20
%28k-2%29%5E2=36%2F29
k-2=±6%2Fsqrt%2829%29
k=2±6%2Fsqrt%2829%29
h=6-%285%2F2%29(2±6%2Fsqrt%2829%29)=6-5±15%2Fsqrt%2829%29=1±15%2Fsqrt%2829%29

So we have two circles with different centers:

(1-15%2Fsqrt%2829%29,2%2B6%2Fsqrt%2829%29) and (1%2B15%2Fsqrt%2829%29,2-6%2Fsqrt%2829%29)
or approximately
(-1.78,3.1) and (3.8, 0.9)

equations:
%28x-%281%2B15%2Fsqrt%2829%29%29%29%5E2%2B%28y-%282%2B6%2Fsqrt%2829%29%29%29%5E2=9
and
%28x-%281-15%2Fsqrt%2829%29%29%29%5E2%2B%28y-%282-6%2Fsqrt%2829%29%29%29%5E2=9



Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Let the center of the circle be (h,k). 
   

The red line is the given line
5x-2y-1 = 0.  The green line is the line perpendicular to it.

Since 5x-2y-1 = 0 solved for y is y=expr%285%2F2%29x-1%2F2, its slope
is 5%2F2 so the green line has slope -2%2F5 and since it
goes through (1,2) we use the point slope formula and get the equation:

y=y%5B1%5D=m%28x-x%5B1%5D%29

y-2=expr%28-2%2F5%29%28x-1%29
5y-10=-2%28x-1%29
5y-10=-2x%2B2
2x%2B5y=12 <-- equation of green line.

Since (h,k) lies on the green line,

2h%2B5k=12

Next we use the formula for the distance from a point (x1,y1) to the line
Ax+By+C=0, which is



and the distance from (h,k) must be ±3 units to the line 5x-2y-1=0, so we have



matrix%281%2C2%2C%22%22+%2B-+abs%285h-2k-1%29%2Fsqrt%2829%29%2C%22%22%29=%22%22+%2B-+3

5h-2k-1+=+%22%22+%2B-+3sqrt%2829%29

5h-2k+=+1+%2B-+3sqrt%2829%29

We solve the system:

system%282h%2B5k=12%2C5h-2k+=+1+%2B-+3sqrt%2829%29%29

using the + and then the -, the two centers are

(h,k) = (-1.785439973,3.114172029)

and

(h,k) = (3.785430073,0.8858279709)

Edwin