Question

The equation of circle having a diameter with endpoints (-2, 1) and (6, 7) is

Answer 1

Hope this helps you!

Answer 2

`(x-2)^2 +(y-4)^2=25`

Explanation:

Since, the diameter of the circle having the end points (-2, 1) and (6, 7),

Thus, by the distance formula,

`\text{Diameter}=√((6+2)^2+(7-1)^2)`

`=√(8^2+6^2)`

`=√(64+36)`

`=√(100)`

`=10\text{ unit}`

So, the radius of the circle =
`(10)/(2)` = 5 unit,

Let the equation of the circle is,

`(x-h)^2+(y-k)^2=25`

Where, (h, k) is the center of the circle,

Since, points (-2, 1) and (6, 7) are on the circle,

They must satisfy the equation of the circle,

`(-2-h)^2+(1-k)^2 = 25`

`(6-h)^2+(7-k)^2=25`

By solving these equations,

We get,

h = 2, k = 4

Hence, the equation of the circle would be,

`(x-2)^2 +(y-4)^2=25`