SOLUTION: Find an equation of the circle with center at(3,-1) that is tangent to the y-axis in the form of {{{(x-A)^2+(y+B)^2=C}}} where A,B,C are constant.Then A= B= C= I already have

Algebra ->  Rational-functions -> SOLUTION: Find an equation of the circle with center at(3,-1) that is tangent to the y-axis in the form of {{{(x-A)^2+(y+B)^2=C}}} where A,B,C are constant.Then A= B= C= I already have       Log On


   

Question 160207: Find an equation of the circle with center at(3,-1) that is tangent to the y-axis in the form of %28x-A%29%5E2%2B%28y%2BB%29%5E2=C where A,B,C are constant.Then
A=
B=
C=
I already have the answers to A and B they are 3 and -1 but I do not know C and I have tried to substitute 3 and -1 for a and b
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

If you'd draw the circle, making it tangent to the y-axis,
and with center (3,-1), you'd easily be able to find the radius,
and then you'd know that C would be the square of the radius.



Now draw in the radius from the center (3,-1) over to the y-axis 
where it's tangent:



You can easily tell that radius is 3 units long, so r=3

So the equation:

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

is

%28x-3%29%5E2%2B%28y%2B1%29%5E2=3%5E2

%28x-3%29%5E2%2B%28y%2B1%29%5E2=9

So now you know that A=3, B=-1 and C=9.

Edwin