Precalculus Examples

${(x - 1)}^{2} + {(y - 4)}^{2} = {(\sqrt{40})}^{2}$

Step 1

Simplify ${(\sqrt{40})}^{2}$ .

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Step 1.1

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 1.1.1

Factor $4$ out of $40$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = {\sqrt{4 (10)}}^{2}$

Step 1.1.2

Rewrite $4$ as $2^{2}$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = {\sqrt{2^{2} \cdot 10}}^{2}$

Step 1.2

Pull terms out from under the radical.

${(x - 1)}^{2} + {(y - 4)}^{2} = {(2 \sqrt{10})}^{2}$

Step 1.3

Simplify the expression.

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Step 1.3.1

Apply the product rule to $2 \sqrt{10}$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = 2^{2} {\sqrt{10}}^{2}$

Step 1.3.2

Raise $2$ to the power of $2$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 {\sqrt{10}}^{2}$

Step 1.4

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

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Step 1.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 {(10^{\frac{1}{2}})}^{2}$

Step 1.4.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 \cdot 10^{\frac{1}{2} \cdot 2}$

Step 1.4.3

Combine $\frac{1}{2}$ and $2$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 \cdot 10^{\frac{2}{2}}$

Step 1.4.4

Cancel the common factor of $2$ .

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Step 1.4.4.1

Cancel the common factor.

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 \cdot 10^{\frac{2}{2}}$

Step 1.4.4.2

Rewrite the expression.

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 \cdot 10^{1}$

Step 1.4.5

Evaluate the exponent.

${(x - 1)}^{2} + {(y - 4)}^{2} = 4 \cdot 10$

Step 1.5

Multiply $4$ by $10$ .

${(x - 1)}^{2} + {(y - 4)}^{2} = 40$

Step 2

This is the form of a circle. Use this form to determine the center and radius of the circle.

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Step 3

Match the values in this circle to those of the standard form. The variable $r$ represents the radius of the circle, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from origin.

$r = 2 \sqrt{10}$

$h = 1$

$k = 4$

Step 4

The center of the circle is found at $(h, k)$ .

Center: $(1, 4)$

Step 5

These values represent the important values for graphing and analyzing a circle.

Center: $(1, 4)$

Radius: $2 \sqrt{10}$

Step 6