Question

i need help asap
Which equation represents a circle with diameter endpoints at (−3,−2) and (1,−4)?


A

(x−1)2+(y−3)2=20


B

(x+1)2+(y+3)2=5


C

(x+1)2+(y+3)2=20


D

(x−1)2+(y−3)2=5

Answer 1

Option C: $(x+1)^2+(y+3)^2=20$ is the equation of the circle.

Explanation:

Given that the endpoints of the circle are at (-3,-2) and (1,-4)

The equation of the circle can be determined using the formula,

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

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

Center:

The center of the circle can be determined using the midpoint formula,

$Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )$

Substituting the coordinates (-3,-2) and (1,-4) in the above formula, we get,

$Center=(\frac{-3+1}{2}, \frac{-2-4}{2} )$

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

$Center=(-1,-3 )$

Thus, the center of the circle is at (-1,-3)

Radius:

The radius of the circle can be determined using the distance formula,

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Substituting the coordinates (-3,-2) and (1,-4) in the above formula, we get,

$r=\sqrt{\left(1+3\right)^{2}+\left(-4+2\right)^{2}}$

$r=\sqrt{\left(4\right)^{2}+\left(-2\right)^{2}}$

$r=\sqrt{\left16+\left4}$

$r=\sqrt{20}$

Thus, the radius of the circle is $\sqrt{20}$

Equation of the circle:

Substituting the center and the radius of the circle in the equation $(x-a)^{2}+(y-b)^{2}=r^{2}$, we get,

$(x+1)^2+(y+3)^2=(\sqrt{20} )^2$

Simplifying, we get,

$(x+1)^2+(y+3)^2=20$

Therefore, the equation of the circle is $(x+1)^2+(y+3)^2=20$

Hence, Option C is the correct answer.