# Determine the equation of the circle whose circumference is 14*pi and the center is (3,-2).

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Given the circumference of a circle is 14pi and the center is at the point (3,-2).

We need to determine the equation of the circle.

First, we need to write into the standard form of the circle:

(x-a)^2 + ( y-b)^2 =r^2 where (a,b) is the center and r is the radius.

==> (x-3)^2 + ( y+2)^2 = r^2.

Now we need to find the radius of the circle.

Given the circumference C = 14pi.

==> 2*r*pi = 14pi

==? r= 14pi/ 2pi = 7

Then the radius of the circle is 7.

==> (x-3)^2 + ( y+2) = 7^2

**==> (x-3)^2 + ( y+2)^2 = 49**

We have to find the equation of the circle whose circumference is 14pi and the center is (3,-2).

Now the circumference of a circle is given by 2*pi*r where r is the radius. Here 2*pi*r = 14*pi

=> r = 14*pi / 2*pi

=> r = 7

Now the equation of a circle with center (h,k) and radius r is (x-h)^2 +(y-k)^2 = r^2

Using the values we have here in the equation we get:

(x- 3)^2 + (y+2)^2 = 49

**Therefore the equation of the circle is (x- 3)^2 + (y+2)^2 = 49.**

The equation of the circle in the standard form is:

(x-h)^2 + (y-k)^2 = r^2

if the circumference of the circle is 14pi, we can compute the radius of the circle:

2*pi*r = 14pi

We'll divide by 2pi:

r = 14pi/2pi

r = 7

We'll identify the coordinates of the circle:

h = 3 and k = -2

**The equation of the circle is:**

**(x-3)^2 + (y+2)^2 = 49**