SOLUTION: What is the equation of the circle passing through (6,2) and tangent to the line x-4y-15=0 at (3,-3).

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Question 1086213: What is the equation of the circle passing through (6,2) and tangent to the line x-4y-15=0 at (3,-3).
Found 3 solutions by htmentor, Edwin McCravy, ikleyn:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
The center of the circle lies on the line perpendicular to the tangent line and passing through (3,-3).
Putting x-4y-15=0 in standard form we have y = x/4 - 15/4
Thus the equation of the line passing through the center is:
y + 3 = -4(x - 3) -> y = -4x + 9
The distance from the center of the circle to each of the two points is equal to the radius of the circle, by definition.
Let the center of the circle be the point (a,b)
Therefore sqrt((a-3)^2 + (b+3)^2) = sqrt((a-6)^2 + (b-2)^2)
And since the center lies on y = -4x + 9, we can substitute for b:
b = -4a + 9
Making the substitution, squaring both sides and collecting terms we are left with:
34a = 68 or a = 2
Therefore b = -4*2 + 9 = 1
So the center is (2,1) and the radius R = sqrt((2-3)^2+(1+3)^2) = sqrt(17)
Thus the equation of the circle is (x-2)^2 + (y-1)^2 = 17

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a different approach that uses calculus, 
in case that's what you're studying. 

Let the equation of the circle be:

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

Substitute the two given points for (x,y)
 
%286-h%29%5E2%2B%282-k%29%5E2=r%5E2
%283-h%29%5E2%2B%28-3-k%29%5E2=r%5E2

Subtracting the two equations:

 %286-h%29%5E2-%283-h%29%5E2%2B%282-k%29%5E2-%28-3-k%29%5E2=0
%286-h%29%5E2-%283-h%29%5E2=%28-3-k%29%5E2-%282-k%29%5E2

Factoring both sides as the difference of squares:



%283%5E%22%22%29%289-2h%5E%22%22%29=%28-5%5E%22%22%29%28-1-2k%5E%22%22%29
27-6h=5%2B10k
22=6h%2B10k
11=3h%2B5k

Here come da calculus:
 
The derivative (slope) of the line must be
the same as the derivative of the circle at 
the point of tangency
 
 x-4y-15=0 is
 1-4expr%28dy%2Fdx%29=0
 -4expr%28dy%2Fdx%29=-1
 dy%2Fdx=1%2F4

Let the circle have the equation

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

Its derivative (by implicit differentiation) is

2%28x-h%29%2B2%28y-k%29expr%28dy%2Fdx%29=0

Divide through by 2

%28x-h%29%2B%28y-k%29expr%28dy%2Fdx%29=0

We substitute the point of tangency (3,-3) and the
derivative = 1/4: 

%283-h%29%2B%28-3-k%29expr%281%2F4%29=0
4%283-h%29%2B%28-3-k%29=0
12-4h-3-k=0
9=4h%2Bk

So we have the system of equations:

system%2811=3h%2B5k%2C9=4h%2Bk%29

Solve the 2nd for k: 9-4h=k.  Substitute in 1st:

 11=3h+5(9-4h)
 11=3h+45-20h
-34=-17h
  2=h
  9=4h+k
  9=4(2)+k
  9=8+k
  1=k

So the center is (h,k) = (2,1)

To find the radius r (all you need for 
the equation is rē), substitute in

%283-h%29%5E2%2B%28-3-k%29%5E2=r%5E2
%283-2%29%5E2%2B%28-3-1%29%5E2=r%5E2
%281%29%5E2%2B%28-4%29%5E2=r%5E2
1%2B16=r%5E2
17=r%5E2 

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

becomes:
 
%28x-2%29%5E2%2B%28y-1%29%5E2=17



Edwin

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
There is the third way to solve this problem.

It is to find the center as the intersection of two straight lines.

One line is the perpendicular to the given line at the tangent point,
and the other line is the perpendicular bisector to the segment, connecting two given points.

In this way you need to solve only one system of two linear equations,
and you avoid solving the quadratic equation.


For many solved problems/samples of this kind see the lesson
    - Find the standard equation of a circle
in this site.