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

Mar 14, 2018

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

#### Explanation:

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

Then, take out the length of the diameter using distance formula,
$\sqrt{{\left(- 2 - 6\right)}^{2} + {\left(2 - \left(- 6\right)\right)}^{2}}$
$\sqrt{64 + 64}$
$\sqrt{128}$
$\sqrt{{2}^{7}}$
$2 \cdot 2 \cdot 2 \sqrt{2}$
$8 \sqrt{2}$
Now, $r a \mathrm{di} u s = \frac{\mathrm{di} a m e t e r}{2}$
radius$= 4 \sqrt{2}$

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

Hope this helps :)
Pretty long one, though.