Algebra Examples

$x^{2} + y^{2} - 2 a x = 0$

Step 1

Since

$x = r \cos (θ)$ , replace

$x$ with

$r \cos (θ)$ .

${(r \cos (θ))}^{2} + y^{2} - 2 a (r \cos (θ)) = 0$

Step 2

Since

$y = r \sin (θ)$ , replace

$y$ with

$r \sin (θ)$ .

${(r \cos (θ))}^{2} + {(r \sin (θ))}^{2} - 2 a (r \cos (θ)) = 0$

Step 3

Solve for

$r$ .

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Step 3.1

Simplify the left side of the equation.

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

Apply the product rule to

$r \cos (θ)$ .

$r^{2} \cos^{2} (θ) + {(r \sin (θ))}^{2} - 2 a (r \cos (θ)) = 0$

Step 3.1.1.2

Apply the product rule to

$r \sin (θ)$ .

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 2 a r \cos (θ) = 0$

Step 3.1.2

Simplify with factoring out.

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Step 3.1.2.1

Factor

$r^{2}$ out of

$r^{2} \cos^{2} (θ)$ .

$r^{2} (\cos^{2} (θ)) + r^{2} \sin^{2} (θ) - 2 a r \cos (θ) = 0$

Step 3.1.2.2

Factor

$r^{2}$ out of

$r^{2} \sin^{2} (θ)$ .

$r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ)) - 2 a r \cos (θ) = 0$

Step 3.1.2.3

Factor

$r^{2}$ out of

$r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ))$ .

$r^{2} (\cos^{2} (θ) + \sin^{2} (θ)) - 2 a r \cos (θ) = 0$

Step 3.1.3

Rearrange terms.

$r^{2} (\sin^{2} (θ) + \cos^{2} (θ)) - 2 a r \cos (θ) = 0$

Step 3.1.4

Apply pythagorean identity.

$r^{2} \cdot 1 - 2 a r \cos (θ) = 0$

Step 3.1.5

Multiply

$r^{2}$ by

$1$ .

$r^{2} - 2 a r \cos (θ) = 0$

Step 3.2

Factor

$r$ out of

$r^{2} - 2 a r \cos (θ)$ .

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Step 3.2.1

Factor

$r$ out of

$r^{2}$ .

$r \cdot r - 2 a r \cos (θ) = 0$

Step 3.2.2

Factor

$r$ out of

$- 2 a r \cos (θ)$ .

$r \cdot r + r (- 2 a \cos (θ)) = 0$

Step 3.2.3

Factor

$r$ out of

$r \cdot r + r (- 2 a \cos (θ))$ .

$r (r - 2 a \cos (θ)) = 0$

Step 3.3

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$r = 0$

$r - 2 a \cos (θ) = 0$

Step 3.4

Set

$r$ equal to

$0$ .

$r = 0$

Step 3.5

Set

$r - 2 a \cos (θ)$ equal to

$0$ and solve for

$r$ .

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Step 3.5.1

Set

$r - 2 a \cos (θ)$ equal to

$0$ .

$r - 2 a \cos (θ) = 0$

Step 3.5.2

Add

$2 a \cos (θ)$ to both sides of the equation.

$r = 2 a \cos (θ)$

Step 3.6

The final solution is all the values that make

$r (r - 2 a \cos (θ)) = 0$ true.

$r = 0$

$r = 2 a \cos (θ)$

$r = 0$

$r = 2 a \cos (θ)$