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

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

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Answer 1 · 2018-03-14T13:48:08

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

Explanation:

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

Then, take out the length of the diameter using distance formula,
$\sqrt{{(- 2 - 6)}^{2} + {(2 - (- 6))}^{2}}$ $sqrt((-2-6)^2+(2-(-6))^2)$
$\sqrt{64 + 64}$ $sqrt(64+64)$
$\sqrt{128}$ $sqrt(128)$
$\sqrt{2^{7}}$ $sqrt(2^7)$
$2 \cdot 2 \cdot 2 \sqrt{2}$ $2*2*2sqrt(2)$
$8 \sqrt{2}$ $8sqrt2$
Now, $r a d i u s = \frac{d i a m e t e r}{2}$ $radius=(diameter)/2$
radius $= 4 \sqrt{2}$ $=4sqrt2$

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

Hope this helps :)
Pretty long one, though.

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

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