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COLLEGE ALGEBRA. LIAL HORNSBY SCHNEIDER. 2.2. Circles. Center-Radius Form General Form An Application. Circle-Radius form.

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## COLLEGE ALGEBRA

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**COLLEGE ALGEBRA**LIAL HORNSBY SCHNEIDER**2.2**Circles Center-Radius Form General Form An Application**Circle-Radius form**By definition, a circle is the set of all points in a plane that lie a given distance from a given point. The given distance is the radius of the circle, and the given point is the center.**Center-Radius Form of the Equation of a Circle**A circle with center (h, k) and radius r has equation which is the center-radius form of the equation of the circle. A circle with center (0, 0) and radius r has equation**FINDING THE CENTER-RADIUS FORM**Example 1 Find the center-radius form of a circle with a center at (–3, 4), radius 6. Solution: a. Use (h, k) = (–3, 4) and r = 6 Center-radius form Substitute Watch signs here.**FINDING THE CENTER-RADIUS FORM**Example 1 Find the center-radius form of a circle with a center at (–3, 4), radius 6. a. Use (h, k) = (–3, 4) and r = 6 Solution Substitute**FINDING THE CENTER-RADIUS FORM**Example 1 b. Find the center-radius form of a circle with a center at (0, 0), radius 3. Solution Because the center is the origin and r = 3, the equation is**GRAPHING CIRCLES**Example 2 Graph the circle. a. Solution Gives (–3, 4) as the center and 6 as the radius.**GRAPHING CIRCLES**Example 2 y Graph the circle. a. Solution 6 (x, y) (–3, 4) x Gives (–3, 4) as the center and 6 as the radius.**GRAPHING CIRCLES**Example 2 y Graph the circle. b. Solution (x, y) 3 x Gives (0, 0) as the center and 3 as the radius.**General Form of the Equation of a Circle**The equation for some real numbers c, d, and e, can have a graph that is a circle or a point, or is nonexistent.**General Form of the Equation of a Circle**Consider There are three possibilities for the graph based on the value of m. • If m > 0, then r2 = m, and the graph of the equation is a circle with the radius • If m = 0, then the graph of the equation is the single point (h, k). • If m < 0, then no points satisfy the equation and the graph is nonexistent.**FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE**Example 3 Show that x2 – 6x + y2 +10y + 25 = 0 has a circle as a graph. Find the center and radius. Solution We complete the square twice, once for x and once for y. and**FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE**Example 3 Add 9 and 25 on the left to complete the two squares, and to compensate, add 9 and 25 on the right. Complete the square. Factor Add 9 and 25 on both sides. Since 9 > 0, the equation represents a circle with center at (3, –5) and radius 3.**FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE**Example 4 Show that 2x2 + 2y2 – 6x +10y = 1 has a circle as a graph. Find the center and radius. Solution To complete the square, the coefficients of the x2- and y2-terms must be 1. Group the terms; factor out 2.**FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE**Example 4 Group the terms; factor out 2. Be careful here.**FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE**Example 4 Factor; simplify on the right. Divide both sides by 2.**FINDING THE CENTER AND RADIUS BY COMPLETING THE SQUARE**Example 4 Divide both sides by 2.**DETERMINING WHETHER A GRAPH IS A POINT OR NONEXISTENT**Example 5 The graph of the equation x2 + 10x + y2 – 4y +33 = 0 is either a point or is nonexistent. Which is it? Solution We complete the square for x and y. Subtract 33. and**DETERMINING WHETHER A GRAPH IS A POINT OR NONEXISTENT**Example 5 The graph of the equation x2 + 10x + y2 – 4y +33 = 0 is either a point or is nonexistent. Which is it? and Complete the square. Factor; add.**DETERMINING WHETHER A GRAPH IS A POINT OR NONEXISTENT**Example 5 Since –4 < 0, there are no ordered pairs (x, y), with both x and y both real numbers, satisfying the equation. The graph of the given equation is nonexistent; it contains no points. ( If the constant on the right side were 0, the graph would consist of the single point (–5, 2).)**LOCATING THE EPICENTER OF AN EARTHQUAKE**Example 6 Three receiving stations are located on a coordinate plane at points (1, 4), (–3, –1), and (5, 2). The distance from the earthquake epicenter to each station should be 2 units, 5 units, and 4 units respectively. Solution Graph the three circles. From the graph it appears that the epicenter is located at (1, 2). To check this algebraically, determine the equation for each circle and substitute x = 1 and y = 2.**LOCATING THE EPICENTER OF AN EARTHQUAKE**Example 6 Station A:**LOCATING THE EPICENTER OF AN EARTHQUAKE**Example 6 Station B:**LOCATING THE EPICENTER OF AN EARTHQUAKE**Example 6 Station C:**LOCATING THE EPICENTER OF AN EARTHQUAKE**Example 6 Three receiving stations are located on a coordinate plane at points (1, 4), (–3, –1), and (5, 2). The distance from the earthquake epicenter to each station should be 2 units, 5 units, and 4 units respectively. The point (1, 2) does lie on all three graphs; thus, we can conclude that the epicenter is at (1, 2).

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