Question

Write the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5).

Answer 1

`(x+1)^2 + (y- 3.5)^2 =64`

Explanation:

We are given that endpoints of a diameter at the points (7,2) and (-9,5).

Center of the circle is the mid point of diameter

So, first find the mid point of diameter

Formula :
`x=(x_1+x_2)/(2) , y=(y_1+y_2)/(2)`

`(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)`

Substitute the values in formula

`x=(7-9)/(2) , y=(2+5)/(2)`

`x=-1 , y=3.5`

So, center of circle = (h,k )=(-1,3.5)

To find length of diameter :

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

`(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)`

`d=√((-9-7)^2+(5-2)^2)`

`d=√(265)`

Length of radius = r =
`(Diameter)/(2)=(√(256))/(2)`

Standard form of the equation of the circle :
`(x -h)^2 + (y- k)^2 = r^2`

(h,k )=(-1,3.5)

`r=(√(256))/(2)`

Equation of the circle :
`(x+1)^2 + (y- 3.5)^2 =((√(256))/(2))^2`

Equation of the circle :
`(x+1)^2 + (y- 3.5)^2 =64`

Hence the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5) is
`(x+1)^2 + (y- 3.5)^2 =64`