Question

For the circle whose diameter has endpoints of (-3, 6) and (5, -2), write the equation in standard form.

Answer 1

Answer:

B

Step-by-step explanation:

We are given a circle whose diameter has endpoints (-3, 6) and (5, -2).

And we want to equation of the circle in standard form.

First, let's determine the center of the circle. Since we are given the diameter, the center will be the midpoint of the diameter. The midpoint is given by:

$\displaystyle M=\Big(\frac{x_1+x_2}{2},\frac{y_1+y_1}{2}\Big)$

By substitution:

$\displaystyle M=\Big(\frac{(-3)+(5)}{2},\frac{(6)+(-2)}{2}\Big)$

Evaluate:

$\displaystyle M=(1, 2)$

Thus, the center of our circle is (1, 2).

Next, we need to find the radius of our circle. We can use the distance formula to find the diameter, and then divide that by two to find the radius. The distance formula is given by:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Let (-3, 6) be (x₁, y₁) and (5, -2) be (x₂, y₂). Substitute:

$d=\sqrt{(-3-5)^2+(6-(-2))^2}$

Evaluate:

$\begin{aligned} d&= \sqrt{(-8)^2+(8)^2}\\&=\sqrt{64+64}\\&=\sqrt{2(64)}\\&=8\sqrt{2}}\end{aligned}$

Therefore, the radius will be:

$\displaystyle r=\frac{8\sqrt{2}}{2}=4\sqrt{2}$

The equation for a circle is given by:

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

Where (h, k) is the center.

By substituting everything in, we acquire:

$(x-(1))^2+(y-(2))^2=(4\sqrt{2})^2$

Simplify:

$(x-1)^2+(y-2)^2=32$

Therefore, our answer is B.