Question

Find a equation of a circle that has a diameter with endpoints -5,-3 and -11,-3

Answer 1

Answer:
`(x+8)^(2)` +
`(y+3)^(2)` =
`3^(2)`

Explanation:

The equation of a circle is given as :

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

Where (a,b) are the coordinate of the center and r is the radius of the circle.

The end point of the diameter is give as (-5, -3 ) and ( -11 , -3 ) , This means that we can find the coordinate of the center by finding the mid point of the end point. The mid point is calculated by :

Mid point = (
`(x_(1)+x_(2))/(2)` ,
`(y_(1)+y_(2))/(2)` )

`x_(1)` = -5

`x_(2)` = -11

`y_(1)` = -3

`y_(2)` = -3

Substituting this values into the formula for finding mid point , we have

Mid point = (
`(-5 - 11)/(2)` ,
`(-3 - 3)/(2)`

Mid-point = (-8 , -3)

Remember that Radius is half of the diameter , To find the diameter we must find the distance between the two end point using the formula for calculating distance between two points , that is

D =
`\sqrt{(x_(2)-x_(1))  ^(2)+(y_(2)-y_(1))  ^(2)}`

Substituting the values :

D =
`\sqrt{(-11+5)^(2)+(-3+3)^(2)}`

D =
`√(36)`

D = 6

Therefore , The diameter i s 6

And since radius is half of the diameter , radius is thus

r = 6/2

r = 3

So , substituting the values gotten into the equation of circle , we have:

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

`(x+8)^(2)` +
`(y+3)^(2)` =
`3^(2)`