# Equation for a circle.

• August 11th 2008, 05:22 PM
MatteNoob
Equation for a circle.
Hello.

I want to rewrite this equation with respect to y

$x^2 + (y-3) = 25$

$y^2 - 6y = 25-9 - x^2$

$y^2 - 6y = 16-x^2$

• August 11th 2008, 05:35 PM
mr fantastic
Quote:

Originally Posted by MatteNoob
Hello.

I want to rewrite this equation with respect to y

$x^2 + {\color{red}(y-3)^2} = 25$

$y^2 - 6y = 25-9 - x^2$

$y^2 - 6y = 16-x^2$

Note the correction in red.
$x^2 + {\color{red}(y-3)^2} = 25$

$\Rightarrow (y-3)^2 = 25 - x^2$

$\Rightarrow y - 3 = \pm \sqrt{25 - x^2}$

$\Rightarrow y = \pm \sqrt{25 - x^2} + 3$
• August 11th 2008, 05:50 PM
MatteNoob
Mr Fantastic

So if I want to write the equation with respect to y, I could do this:

$(x-3)^2 +(y+1)^2 = 25$

$(y+1)^2 = 5^2 - (x-3)^2$

$y = \pm \sqrt{5^2 - (x-3)^2}-1$

Then, if I want the derivative, I could;

$u = -x^2 + 6x -16 \;\;\;\; u\prime = -2x +6$

$y\prime = \pm \left(\sqrt{u}\right)\prime \cdot u\prime - (1)\prime$

So that

$y\prime = \pm \frac{-2x+6}{2\sqrt{-x^2+6x-16}} = \underline{\underline{ \pm \frac{3-x}{\sqrt{6x - x^2-16}}}}$

Edit: Corrected a typo.
• August 11th 2008, 09:00 PM
mr fantastic
Quote:

Originally Posted by MatteNoob
Mr Fantastic

So if I want to write the equation with respect to y, I could do this:

$(x-3)^2 +(y+1)^2 = 25$

$(y+1)^2 = 5^2 - (x-3)^2$

$y = \pm \sqrt{5^2 - (x-3)^2}-1$

Then, if I want the derivative, I could;

$u = -x^2 + 6x {\color{red}+} 16 \;\;\;\; u\prime = -2x +6$

$y\prime = \pm \left(\sqrt{u}\right)\prime \cdot u\prime - (1)\prime$

So that

$y\prime = \pm \frac{-2x+6}{2\sqrt{-x^2+6x {\color{red}+} 16}} = \underline{\underline{ \pm \frac{3-x}{\sqrt{6x - x^2 {\color{red}+} 16}}}}$

Edit: Corrected a typo.

Note the corrections in red.