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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 11: Q. 42 (page 1083)

In the following exercises, 鈸� identify the center and radius, and 鈸� graph.
$x^{2} + y^{2} - 6 x - 8 y = 0$

Short Answer

Expert verified

(a) Center of the circle $x^{2} + y^{2} - 6 x - 8 y = 0$ is $(3, 4)$ and radius is $5$

(b) graph of the circle is

Step by step solution

01

Step 1. given data

The given equation of the circle is

$x^{2} + y^{2} - 6 x - 8 y = 0$

02

Step 2. Part (a)

Group the x-terms and y-terms

$x^{2} + y^{2} - 6 x - 8 y = 0 (x^{2} - 6 x) + (y^{2} - 8 y) = 0$

complete the squares

$(x^{2} - 6 x + 9) + (y^{2} - 8 y + 16) = 9 + 16 {(x - 3)}^{2} + {(y - 4)}^{2} = 5^{2}$

Compare the equation with the standard equation of the circle

${(x - 3)}^{2} + {(y - 4)}^{2} = 5^{2} {(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

so center $(h, k) = (3, 4)$

radius $r = 5$

03

Step 3. Part (b)

Plot the circle of center $(3, 4)$ and radius $5$

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Most popular questions from this chapter

In the following exercises, factor each trinomial of the form

$x^{2} + b x y + c y^{2} m^{2} - 64 m n - 65 n^{2}$

Simplify:

(a) ${(- 256)}^{\frac{1}{4}}$

(b) ${- 256}^{\frac{1}{4}}$

(c) ${(256)}^{- \frac{1}{4}}$

In the following exercises, simplify each expression.

${(y^{4})}^{3} . {(y^{5})}^{2}$

In the following exercises, factor completely using the difference of squares pattern, if possible.

$a^{2} + 6 a + 9 - 9 b^{2}$

Factor $a^{2} - 11 a b + 10 b^{2}$

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