How do you find an equation of the circle that satisfies the given conditions: endpoints of a diameter are P(−1, 2) and Q(7, 8)?

The coordinates of the centre of the circle will be the coordinates of the midpoint of the diameter.

`((-1+7)/2,(2+8)/2)=(3,5)`

We next find the length of the diameter using the distance formula:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`d=sqrt((7-(-1))^2+(8-2)^2)=sqrt(100)=10`

This is the length of the diameter, so radius is:

`r=10/2=5`

The equation of a circle is given as:

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

Where `h and k` are the `x and y` coordinates of the centre respectively.

`:.`

`(x-3)^2+(y-5)^2=25`

Equation of a circle is of the form: `(x-a)^2 + (y-b)^2 = r^2`
Where (a,b) represents the co ordinates of the circle's centre
And 'r', the circle's radius
Equation to find the coordinates of the centre of a line segment is
= `((x1 + x2)/2) , ((y1 + y2)/2)`
Hence,
( 0.5 ( -1 + 7 ) ) , ( 0.5 ( 2 + 8 ) )
= ( 3 , 5 )
For the distance between two points , `sqrt(( x2 - x1 )^2 + ( y2 - y1 )^2)`
Hence, `sqrt( ( 7 - ( -1 ) )^2 + ( 8 - 2 )^2`
= 10
Diameter of circle = 10
Radius , r = 5
= `( x - 3 )^2 + ( y - 5 )^2 = 5^2`
= `( x - 3 )^2 + ( y- 5 )^2 = 25`