Question

Write the standard form of equation of a circle having a diameter whose endpoints are ( 3,-2 ) & ( 2,9 )?

Answer 1

`color(blue)((x-5/2)^2+(y-7/2)^2=61/2)`

Explanation:

The standard form of the equation of a circle is given as:

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

Where:

`bbh` and `bbk` are the the `bbx` and `bby` co-ordinates of the centre respectively, and `bbr` is the radius.

We have the endpoints of the diameter and we know that the diameter always passes through the centre of a circle. The co-ordinates of the centre will be the co-ordinates of the midpoint of the diameter.

We first find the length of the diameter, since half of this will be the radius.

Using the distance formula:

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

Endpoints:

`(3,-2) ,(2,9)`

`:.`

`|d|=sqrt((2-3)^2+(9-(-2))^2)=sqrt(122)`

So our radius is:

`r=(sqrt(122))/2`

`r^2=122/4=61/2`

The co-ordinates of the midpoint of a line segment is given as:

`"midpoint"=((x_1+x_2)/2, (y_1+y_2)/2)`

Using: `(3,-2) ,(2,9)`

`((3+2)/2,(-2+9)/2)=(5/2,7/2)=>h=5/2 and k=7/2`

We can now find the equation:

`(x-5/2)^2+(y-7/2)^2=61/2`

PLOT: