Given the equation of a circle above, find the center and the radius.?

Question

http://i40.photobucket.com/albums/e215/xxs3xp1stolxx/Picture5-1.png

Answer 1

(x - 5)² + (y + 3)² = 16
(x - 5)² + (y - (- 3))² = 4²
Centre (5,- 3)
Radius 4

Answer 2

to discover the midsection of a circle, you do the different of the style interior the parenthesis. because of the fact the equation if (x-5)² + (x+3)² = sixteen. the midsection would be (5, -3). And the radius is got here across by utilising sq. rooting the style to the appropriate of the equivalent sign. So it would be 4.

Answer 3

When finding the center, you take the oposite of what is inside the parentheses. Since it is x minus 5, the center lies at x=postive 5. Likewise for y: since it is y plus three, the center lies at y minus three.

To find the radius, it is simply the square root of what the equation is equal to. So the square root of 16=4.

So the center lies at (5,-3) with a radius of 4.

Answer 4

The right answer is the last one:

Center: (5 , -3)
Radius = √(16) = 4

Answer 5

the general equation of a circle is:

(x-a)^2+(y-b)^2=r^2

where the circle is centered at (a,b) and has radius r

so, this circle is centered at (5,-3) and has a radius of 4

Answer 6

comparing to (x-h)^2+(y-k)^2=r^2
-h=-5
h=5
-k=3
k=-3
r^2=16
r=4
centre=(h,k)=(5,-3)
radius=4 unit
the answer is the fourth in the image.

Answer 7

its the last one. Previously Answered.