# How do you write an equation in standard form of the circle with the given properties: endpoints of a diameter are (9,2) and (-9,-12)?

Apr 24, 2016

$\implies {\left(y + 5\right)}^{2} + {x}^{2} = {r}^{2}$

#### Explanation:

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The diameter length will be the distance between (9,2) and (-9,-12)

$\implies 2 r = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$\implies 2 r = \sqrt{{\left(- 9 - 9\right)}^{2} + {\left(- 12 - 2\right)}^{2}} = \sqrt{520}$

$\implies 2 r = 2 \sqrt{130} \text{ "=>" } r = \sqrt{130}$
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Known: The equation of a circle where the centre is coincidental to the origin is ${y}^{2} + {x}^{2} = {r}^{2}$

But the centre of the circle does not coincide with the origin.
So the equation needs to be adjusted. We have to make allowance for offsetting both the x and y.

Centre of the circle is at the mean x and the mean y

Centre$\to \left(x , y\right) \to \left(\frac{9 - 9}{2} , \frac{2 - 12}{2}\right) = \left(0 , - 5\right)$

So we need to 'translate' (mathematically move) the coordinates of the circles centre back to the origin to relate it to ${r}^{2.}$

$\implies {\left(y + 5\right)}^{2} + {\left(x + 0\right)}^{2} = {r}^{2}$

$\implies {\left(y + 5\right)}^{2} + {x}^{2} = {r}^{2}$

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

Consider ${\left(y + 5\right)}^{2} + {x}^{2} = {r}^{2}$

$\implies y = \pm \sqrt{{r}^{2} - {x}^{2}} \textcolor{w h i t e}{. .} - 5$

Please ignore the red dotted lines. My package is doing something odd!