Question

If the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?

Answer 1

1. Find the centre of the circle or the midpoint of the two numbers
(6-6/2}, (-2+6/2)
(0, 2)
2. Find radius using the distance formula
√(6 - 2)² + (-6 - 0)²
√4² + (-6)²
√16 + 36
√52
r = 7.21
3. Plug in
x² + (y - 2)² = 52

So it is option D 99% sure

Answer 2

Answer:

(D) $(x-0)^2+(y-2)^2=52$

Step-by-step explanation:

It is given that the endpoints of the diameter of a circle are (−6, 6) and (6, −2).

Now, the standard form equation of the circle is:

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

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

In order to find the center, first find the midpoint of the two given points, that is:

$C=(\frac{-6+6}{2}, \frac{6-2}{2})$

$c=(0,2)$

Thus, the center is (0,2).

Now, the radius is the distance from the center to either of  the two given points, therefore using distance formula,

$r^2=(-6-0)^2+(6-2)^2$

$r^2=52$

Also, the equation of the circle is:

$(x-0)^2+(y-2)^2=52$

Hence, option D is correct.