Question

Which equation has a center at (2, 1) and a point on the circle of (2, -3)?

O (x + 2)2 + (y + 1)2 = 16
O x - 2)2 + (y - 1)2 = 4
O(x - 2)2 + (y - 1)2 = 16
O (x + 2)2 + (y + 1)2 = 4

Answer 1

c on edge

i got it right

Answer 2

Option 3:
`(x-2)^2+(y-1)^2 = 16` is the correct answer

Explanation:

We will find the equation of the circle with given information and then compare with the choices given.

Given

Center = (h,k) = (2,1)

And point on circle = (2,-3)

The equation of circle is given as:

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

The distance between center and point on circle is the radius. So using the distance formula:

`r = √((x_2-x_1)^2+(y_2-y_1)^2)\\r = √((2-2)^2+(-3-1)^2)\\r = √((0)^2+(-4)^2)\\r = √(0+16)\\r = √(16)\\or\\r = 4`

Putting the values in equation of circle

`(x-2)^2+(y-1)^2 = 4^2\\(x-2)^2+(y-1)^2 = 16`

Hence,

Option 3:
`(x-2)^2+(y-1)^2 = 16` is the correct answer