Precalculus Examples

(- 4, 3)

$(- 4, 3)$

$r = \sqrt{7}$

Step 1

The standard form of a circle is

$x^{2}$ plus

$y^{2}$ equals the radius squared

$r^{2}$ . The horizontal

$h$ and vertical

$k$ translations represent the center of the circle. The formula is derived from the distance formula where the distance between the center and every point on the circle is equal to the length of the radius.

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

Step 2

Fill in the values of

$h$ and

$k$ which represent the center of the circle.

${(x + 4)}^{2} + {(y - 3)}^{2} = r^{2}$

Step 3

Fill in the value of

$r$ which represents the radius of the circle.

${(x + 4)}^{2} + {(y - 3)}^{2} = {(\sqrt{7})}^{2}$

Step 4

Simplify.

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Step 4.1

Remove parentheses.

${(x + 4)}^{2} + {(y - 3)}^{2} = {\sqrt{7}}^{2}$

Step 4.2

Rewrite

${\sqrt{7}}^{2}$ as

$7$ .

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Step 4.2.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{7}$ as

$7^{\frac{1}{2}}$ .

${(x + 4)}^{2} + {(y - 3)}^{2} = {(7^{\frac{1}{2}})}^{2}$

Step 4.2.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

${(x + 4)}^{2} + {(y - 3)}^{2} = 7^{\frac{1}{2} \cdot 2}$

Step 4.2.3

Combine

$\frac{1}{2}$ and

$2$ .

${(x + 4)}^{2} + {(y - 3)}^{2} = 7^{\frac{2}{2}}$

Step 4.2.4

Cancel the common factor of

$2$ .

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Step 4.2.4.1

Cancel the common factor.

${(x + 4)}^{2} + {(y - 3)}^{2} = 7^{\frac{2}{2}}$

Step 4.2.4.2

Rewrite the expression.

${(x + 4)}^{2} + {(y - 3)}^{2} = 7^{1}$

Step 4.2.5

Evaluate the exponent.

${(x + 4)}^{2} + {(y - 3)}^{2} = 7$

Step 5

$C (- 4, 3) r = \sqrt[]{7}$

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