Holt Algebra Circles 10-2 Circles Holt Algebra2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz

Holt Algebra Circles Warm Up Find the slope of the line that connects each pair of points. 1. (5, 7) and (–1, 6) 2. (3, –4) and (–4, 3)

Holt Algebra Circles Warm Up Find the distance between each pair of points. 3. (–2, 12) and (6, –3) 4. (1, 5) and (4, 1)

Holt Algebra Circles Write an equation for a circle. Graph a circle, and identify its center and radius. Objectives

Holt Algebra Circles circle tangent Vocabulary

Holt Algebra Circles A circle is the set of points in a plane that are a fixed distance, called the radius, from a fixed point, called the center. Because all of the points on a circle are the same distance from the center of the circle, you can use the Distance Formula to find the equation of a circle.

Holt Algebra Circles Write the equation of a circle with center (–3, 4) and radius r = 6. Example 1: Using the Distance Formula to Write the Equation of a Circle Use the Distance Formula with (x 2, y 2 ) = (x, y), (x 1, y 1 ) = (–3, 4), and distance equal to the radius, 6. Use the Distance Formula. Substitute. Square both sides.

Holt Algebra Circles Write the equation of a circle with center (4, 2) and radius r = 7. Check It Out! Example 1

Holt Algebra Circles Notice that r 2 and the center are visible in the equation of a circle. This leads to a general formula for a circle with center (h, k) and radius r. If the center of the circle is at the origin, the equation simplifies to x 2 + y 2 = r 2. Helpful Hint

Holt Algebra Circles Write the equation of the circle. Example 2A: Writing the Equation of a Circle (x – 0) 2 + (y – 6) 2 = 1 2 x 2 + (y – 6) 2 = 1 the circle with center (0, 6) and radius r = 1 (x – h) 2 + (y – k) 2 = r 2 Equation of a circle Substitute.

Holt Algebra Circles Use the Distance Formula to find the radius. Substitute the values into the equation of a circle. (x + 4) 2 + (y – 11) 2 = 225 the circle with center (–4, 11) and containing the point (5, –1) (x + 4) 2 + (y – 11) 2 = 15 2 Write the equation of the circle. Example 2B: Writing the Equation of a Circle

Holt Algebra Circles Find the equation of the circle with center (–3, 5) and containing the point (9, 10). Check It Out! Example 2

Holt Algebra Circles The location of points in relation to a circle can be described by inequalities. The points inside the circle satisfy the inequality (x – h) 2 + (y – k) 2 < r 2. The points outside the circle satisfy the inequality (x – h) 2 + (y – k) 2 > r 2.

Holt Algebra Circles Use the map and information given in Example 3 on page 730. Which homes are within 4 miles of a restaurant located at (–1, 1)? Example 3: Consumer Application The circle has a center (–1, 1) and radius 4. The points insides the circle will satisfy the inequality (x + 1) 2 + (y – 1) 2 < 4 2. Points B, C, D and E are within a 4-mile radius. Point F (–2, –3) is not inside the circle. Check Point F(–2, –3) is near the boundary. (–2 + 1) 2 + (–3 – 1) 2 < 4 2 (–1) 2 + (–4) 2 < < 16 x

Holt Algebra Circles What if…? Which homes are within a 3-mile radius of a restaurant located at (2, –1)? Check It Out! Example 3

Holt Algebra Circles A tangent is a line in the same plane as the circle that intersects the circle at exactly one point. Recall from geometry that a tangent to a circle is perpendicular to the radius at the point of tangency. To review linear functions, see Lesson 2-4. Remember!

Holt Algebra Circles Write the equation of the line tangent to the circle x 2 + y 2 = 29 at the point (2, 5). Example 4: Writing the Equation of a Tangent Step 1 Identify the center and radius of the circle. From the equation x 2 + y 2 = 29, the circle has center of (0, 0) and radius r =.

Holt Algebra Circles Example 4 Continued Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent. Substitute (2, 5) for (x 2, y 2 ) and (0, 0) for (x 1, y 1 ). Use the slope formula. The slope of the radius is. 5 2 Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is. 2 5 –

Holt Algebra Circles Example 4 Continued Use the point-slope formula. Rewrite in slope-intercept form. Substitute (2, 5) (x 1, y 1 ) and – for m. 2 5 Step 3 Find the slope-intercept equation of the tangent by using the point (2, 5) and the slope m =. 2 5 –

Holt Algebra Circles Example 4 Continued The equation of the line that is tangent to x 2 + y 2 = 29 at (2, 5) is. Check Graph the circle and the line.

Holt Algebra Circles Write the equation of the line that is tangent to the circle 25 = (x – 1) 2 + (y + 2) 2, at the point (5, –5). Check It Out! Example 4 Check Graph the circle and the line.