# Write the equation of the circle that has the center (-3,4) and radius 11.

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### 2 Answers

First we will write the standard for for the circle equation.

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

Given that (-3,4) is the center and 11 is the radius.

==> (x +3)^2 + (y-4)^2 = 11^2

Now we will open brackets.

==> x^2 +6x + 9 + y^2 - 8y + 16 = 121

===> x^2 + y^2 + 6x -8y = 121 -16 -9

==> x^2 + y^2 + 6x -8y = 96

Then the equation of the circle is :

**x^2 + y^2 + 6x - 8y = 96**

The standard equation of the circle is:

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

The coordinates h and k represent the coordinates of the center of the circle.

We'll replace h and k by the values of coordinates of the center in the equation of the circle:

(x - (-3))^2 + (y - 4)^2 = 11^2

(x + 3)^2 + (y - 4)^2 = 121

**The equation of the circle whose center is C(-3,4) and radius , r = 11, is: (x + 3)^2 + (y - 4)^2 = 121.**