Calculus Examples

x^{2} + y^{2} = r^{2}

$x^{2} + y^{2} = r^{2}$

Step 1

Subtract

x^{2}

$x^{2}$ from both sides of the equation.

y^{2} = r^{2} - x^{2}

$y^{2} = r^{2} - x^{2}$

Step 2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

y = \pm \sqrt{r^{2} - x^{2}}

$y = \pm \sqrt{r^{2} - x^{2}}$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = r

$a = r$ and

b = x

$b = x$ .

y = \pm \sqrt{(r + x) (r - x)}

$y = \pm \sqrt{(r + x) (r - x)}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

y = \sqrt{(r + x) (r - x)}

$y = \sqrt{(r + x) (r - x)}$

Step 4.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

y = - \sqrt{(r + x) (r - x)}

$y = - \sqrt{(r + x) (r - x)}$

Step 4.3

The complete solution is the result of both the positive and negative portions of the solution.

y = \sqrt{(r + x) (r - x)}

$y = \sqrt{(r + x) (r - x)}$

y = - \sqrt{(r + x) (r - x)}

$y = - \sqrt{(r + x) (r - x)}$

y = \sqrt{(r + x) (r - x)}

$y = \sqrt{(r + x) (r - x)}$

y = - \sqrt{(r + x) (r - x)}

$y = - \sqrt{(r + x) (r - x)}$

x^{2} + y^{2} = r^{2}

$x^{2} + y^{2} = r^{2}$

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