Question

Geometryyyyyyyy!!!!!!!!!!!!!!!

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

B
`(x+1)^(2)` +
`(y+3)^(2)` = 5

Explanation:

Given the diameter endpoints of circle: (-3,-2) and (1, -4)

We know that the equation of circle is

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

where (x,y) is any point on the circle, (h,k) is center of the circle and r is radius of circle.

To find (h,k): the center is midpoint of diameter

Midpoint of diameter with end points (x1,y1) and (x2,y2) is given by

(
`(x1+x2)/(2)` ,
`(y1+y2)/(2)` )

(
`(-3+1)/(2)` ,
`(-2-4)/(2)` )

(-1,-3)

Hence (h,k) is (-1,-3)

Substituting values of (h.k) and (x.y) as (-1,-3) and (1,-4) respectively in equation of circle, we get

`(1+1)^(2)` +
`(-4+3)^(2)` =
`r^(2)`

`r^(2)` = 5

Substituting the values of (h,k) and
`r^(2)`, we get the equation of circle as

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

Hence the answer is B