Question

How do you write an equation for a circle with diameter are P(−2, 2) and Q(6, −6)?

Answer 1

I guess you meant end points of the diameter are `P(-2,2)` and `Q(6,-6)`.

Explanation:

First, you take out the midpoint of the line joining points `P` and `Q`.
Let it be `O`.
`O=((-2+6)/2,(2-6)/2)`
( Using midpoint formula)
`O=(4/2,-4/2)`
`O=(2,-2)`
Since, the midpoint of a diameter is the centre of the circle, and standard coordinate of the centre is `O(h,k)`,
We get `h=2` and `k=-2`.

Then, take out the length of the diameter using distance formula,
`sqrt((-2-6)^2+(2-(-6))^2)`
`sqrt(64+64)`
`sqrt(128)`
`sqrt(2^7)`
`2*2*2sqrt(2)`
`8sqrt2`
Now, `radius=(diameter)/2`
radius`=4sqrt2`

Now, we have `h=2`, `k=-2` and `r=4sqrt2`
Let us substitute these in the standard equation of a circle,
`(h-x)^2+(k-y)^2=r^2`
`(2-x)^2+(-2-y)^2=(4sqrt2)^2`
`4-4x+x^2+4+4y+y^2=32`
`x^2+y^2-4x+4y=24`
This is our required equation.

Hope this helps :)
Pretty long one, though.