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)?

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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)?

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Answer 1 · 2016-04-24T20:27:01

$\Rightarrow {(y + 5)}^{2} + x^{2} = r^{2}$ $=>(y+5)^2+x^2=r^2$

Explanation:

$I automatically did this bit but decided it is not needed.$ $color(blue)("I automatically did this bit but decided it is not needed.")$ $However, I left it in as a demonstration of method$ $color(blue)("However, I left it in as a demonstration of method")$

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

$\Rightarrow 2 r = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $=>2r=sqrt( (x_2-x_1)^2+(y_2-y_1)^2)$

$\Rightarrow 2 r = \sqrt{{(- 9 - 9)}^{2} + {(- 12 - 2)}^{2}} = \sqrt{520}$ $=>2r=sqrt((-9-9)^2+(-12-2)^2) = sqrt(520)$

$\Rightarrow 2 r = 2 \sqrt{130} \Rightarrow r = \sqrt{130}$ $=>2r=2sqrt(130)" "=>" "r=sqrt(130)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$This is the relevant calculation$ $color(blue)("This is the relevant calculation")$

Known: The equation of a circle where the centre is coincidental to the origin is $y^{2} + x^{2} = r^{2}$ $y^2+x^2=r^2$

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 $\to (x, y) \to (\frac{9 - 9}{2}, \frac{2 - 12}{2}) = (0, - 5)$ $->(x,y)->((9-9)/2,(2-12)/2) = (0,-5)$

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

$\Rightarrow {(y + 5)}^{2} + {(x + 0)}^{2} = r^{2}$ $=>(y+5)^2+(x+0)^2=r^2$

$\Rightarrow {(y + 5)}^{2} + x^{2} = r^{2}$ $=>(y+5)^2+x^2=r^2$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")$

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

$\Rightarrow y = \pm \sqrt{r^{2} - x^{2}} . . - 5$ $=>y=+- sqrt (r^2-x^2)color(white)(..) -5$

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

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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)?

1 Answer

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