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

Answer:

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

Step-by-step 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:

$r = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\r = \sqrt{(2-2)^2+(-3-1)^2}\\r = \sqrt{(0)^2+(-4)^2}\\r = \sqrt{0+16}\\r = \sqrt{16}\\or\\r = 4$

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

Answer 2

Answer:

c on edge

i got it right