multiple choice hw problem.......need help!?

Question

The equation of the circle with center at (5, 6) and radius of 7 is ____

CHOICES:
x2 - y2 = 49
(x - 5)2 + (y - 6)2 = 49
(x - 5)2 - (y - 6)2 = 49
(x + 5)2 + (y + 6)2 = 49

Answer 1

choice 2
(x-5)^2+(y-6)^2=49

Answer 2

A circle at the origin (0, 0) of radius 7 has the equation x^2 + y^2 = 7^2.

Your problem needs to move (5, 6) to the origin. You do this just be subtracting the values you want from x and y.

(x - 5)^2 + (y - 6)^2 = 7^2.

Notice if you try (5, 6) in the above equation you get 0^2 + 0^2, which is just what you would expect at the center of a circle.

Answer 3

Start plugging in choices to see if any work

5^2-6^ = 49
25 - 36 does not equal 49

(5-5)^2 + (6-6)^2 = 49
0+0 does not equal 49

The next equation will result in exactly the same as the one I just did

(x+5)^2 + (y+6)^2
(5+5)^2 + (6+6)^2
10^2 + 36^2
100+36 does not equal 49, so none of these are correct.

Answer 4

(x - 5)2 + (y - 6)2 = 49 is correct

Answer 5

Choice 2

Answer 6

One way you could easily solve this is by testing some points you know have to exist on that circle. For example (-2,6) or (5,-1). Plug these values into the left hand equation. If they equal 49, you have your circle equation.

Answer 7

The formula for a circle is:

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

Where (a,b) are the coordinates of the center and r is the radius.

Plug in what you were given and see what you get.

Answer 8

Choice 2 is the right answer

Answer 9

If we have centre as (h,k) and radius as r
then equation is (x-h)^2+(y-k)^2=r^2
so now we know centre is (5,6) and r=7
therefore
(x-5)^2+(y-6)^2=49

Answer 10

I thought a squared plus b squared equals c squared was to find the hypotenuse of a triangle ... but choice D #4 looks right ..... I donno for sure.

Isn't there an equation in your text book that you just plug the numbers in?? That's why Iiiiiii usually succeeded so well in math in HS. :o)