# State the center and radius..Given the equation of a circle, (x+2)^2+(y-1)^2=36, state the center and radius.

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### 3 Answers

The standard equation of a circle with center ( a , b) and a radius of r is given as: (x - a)^2 + (y-b)^2 = r^2

Now we have the equation of the circle of (x+2)^2+(y-1)^2=36

Comparing this with the standard form (x-a)^2 + (y-b)^2 = r^2, we see

a = -2 , b = 1 , r = 6

**Therefore the radius is 6 and the center is (-2 , 1).**

The equation of circle is given the (x-h)^2+(y-h)^2 = r^2...(1), where (h,k) is the coordinates of the centre. and r is the radius of the circle.

The given equation is (x+2)^2 +(y-1)^2 = 36....(2)

If (1) and (2) are representing the same circle, then the like terms , we get :

(x-h)^2 = (x+2)^2.

-2xh = 4x and +h^2 = 2^2 . Therefore h = 4x/-2x = -2.

Similarly, ((y-k)^2 = (y-1)^2 gives: k= 1.

r^2 = 36 , r = (36)^(1/2) = 6.

Therefore the centre (h, k) = (-2,1).

radius r = 6.

So the centre of the circle (x+2)^2+(y-1) = 36 is at (-2,1). The radius is 6.

We'll write the equation of the circle:

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

C(h , k) - the x and y coordinates of the center of the circle

r - the radius of the circle

Comparing the given equation and the equation of the circle, with perfect squares, we'll identify the followings:

(x+2)^2+(y-1)^2=36

h = -2

k = 1

r^2 = 36

r = 6

**The circle has the coordinates of the center C(-2 , 1) and the radius r = 6.**