Which of the following points represents the center of a circle whose equation is (x-3)^2 + (y-2)^2 =16

it’s (3,2)

Step-by-step explanation:

(3, 6)

Explanation:

Circle Function: (x - h)² + (y - k)² = r²

(h, k) of the function is the center, so (3, 6) is the center.

Alternatively, you can graph it and locate the center.

A. (3,2)

Step-by-step explanation:

To find the center of a circle on a quadratic equation all you have to do is to see what´s inside the parenthesis with the X and Y and then you equal that to "0", the number inside the parenthesis is the distance from the center of the circle to the center of the graph.

1. (x-3)= 0

2. x=3

1.  (y-2)

2. Y=2

This means that the center of the circle is 3 units away on the positive x axis and the  Y is 2 units away on the positive Y axis.

B

Step-by-step explanation:

(x-3)² + (y-2)² = 16

(x-h)²+(y-k)²=16

r=4

h=3

k=2

center is at the point (h, k) ⇒ (3, 2)

C

Step-by-step explanation:

The answer is whatever turns the two terms on the left to zero.

x - 3 = 0

x = 3

y - 6 = 0

y = 6

So the answer is (3,6)

which is C

compare with the equation of the circle (x-a)^2 + (y -b )^2 = r^2

(a,b) are the centers of the circle

therefore, the centers are (5,4)

Step-by-step explanation:

D

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r the radius

(x - 5)² + (y - 4)² = 25 ← is in standard form

with centre = (5, 4) → D

The answer is option C

(3 , 6)

Hope this helps.

I think this is geometry. So you would have to use the standard equation of a circle. 
The standard equation of a circle with center (h,k) and a radius of r is: (x-h)^2 + (y-k)^2 = r^2
With that equation, we know that the center of YOUR equation is (3 , 2). 
So now make (x-3) and (y-2) equal to 0. 
That would make x=3 and y=2. 
Therefore the answer is A. 

Your answer is A (3, 2)