holt algebra 2 10-2 circles write an equation for a circle. graph a circle, and identify its center...
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Holt Algebra 2
10-2 Circles
Write an equation for a circle. Graph a circle, and identify its center and radius.
Objectives
circle tangent
Vocabulary
Holt Algebra 2
10-2 CirclesNotes
1. Write an equation for the circle with center (1, –5)
and a radius of .
2. Write an equation for the circle with center (–4, 4) and containing the point (–1, 16).
3. Write an equation for the line tangent to the circle x2 + y2 = 17 at the point (4, 1).
4. Which points on the graph shown are within 2 units of the
point (0, –2.5)?
Holt Algebra 2
10-2 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 2
10-2 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 (x2, y2) = (x, y), (x1, y1) = (–3, 4), and distance equal to the radius, 6.
Use the Distance Formula.
Substitute.
Square both sides.
Holt Algebra 2
10-2 CirclesNotice that r2 and the center are visible in the equation of a circle. This leads to the 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 x2 + y2 = r2.
Helpful Hint
Holt Algebra 2
10-2 Circles
Write the equation of the circle.
Example 2A: Writing the Equation of a Circle
(x – 0)2 + (y – 6)2 = 12
x2 + (y – 6)2 = 1
the circle with center (0, 6) and radius r = 1
(x – h)2 + (y – k)2 = r2 Equation of a circle
Substitute.
Holt Algebra 2
10-2 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 = 152
Write the equation of the circle.
Example 2B: Writing the Equation of a Circle
Holt Algebra 2
10-2 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 + (x – k)2 < r2. The points outside the circle satisfy the inequality (x – h)2 + (x – k)2 > r2.
Holt Algebra 2
10-2 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 < 42. 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 < 42
(–1)2 + (–4)2 < 42
1 + 16 < 16 x
Holt Algebra 2
10-2 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.
Holt Algebra 2
10-2 Circles
Write the equation of the line tangent to the circle x2 + y2 = 29 at the point (2, 5).
Example 4: Writing the Equation of a Tangent
Step 1 Identify the center and radius of the circle.
The circle has center of (0, 0) and radius r = .
Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent.
Use the slope formula.
Holt Algebra 2
10-2 CirclesExample 4 Continued
Use the point-slope formula.
Rewrite in slope-intercept form.
Substitute (2, 5) (x1 , y1 ) 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 = . Perpendicular lines have slopes that are
opposite reciprocals
2 5
–
Holt Algebra 2
10-2 Circles
Example 4 Continued
The equation of the line that is tangent to
x2 + y2 = 29 at (2, 5) is .
Graph the circle and
the line to check.
Holt Algebra 2
10-2 CirclesNotes
1. Write an equation for the circle with center (1, –5) and a radius of .
(x – 1)2 + (y + 5)2 = 102. Write an equation for the circle with center
(–4, 4) and containing the point (–1, 16).
(x + 4)2 + (y – 4)2 = 153
3. Write an equation for the line tangent to the circle x2 + y2 = 17 at the point (4, 1).
y – 1 = –4(x – 4)
Holt Algebra 2
10-2 Circles
Notes
4. Which points on the graph shown are within 2 units of the point (0, –2.5)?
C, F
Holt Algebra 2
10-2 CirclesCircles: Extra Info
The following power-point slides contain extra examples and information.
Reminder of Lesson ObjectivesWrite an equation for a circle. Graph a circle, and identify its center and radius.
Holt Algebra 2
10-2 Circles
Write an equation for a circle. Graph a circle, and identify its center and radius.
circle tangent
Vocabulary
Holt Algebra 2
10-2 Circles
Use the Distance Formula to find the radius.
Substitute the values into the equation of a circle.
(x + 3)2 + (y – 5)2 = 169
Find the equation of the circle with center (–3, 5) and containing the point (9, 10).
(x + 3)2 + (y – 5)2 = 132
Check It Out! Example 2
Holt Algebra 2
10-2 Circles
Write the equation of the line that is tangent to the circle 25 = (x – 1)2 + (y + 2)2, at the point (1, –2).
Step 1 Identify the center and radius of the circle.
Check It Out! Example 4
From the equation 25 = (x – 1)2 +(y + 2)2, the circle has center of (1, –2) and radius r = 5.
Holt Algebra 2
10-2 Circles
Step 2 Find the slope of the radius at the point of tangency and the slope of the tangent.
Substitute (5, –5) for (x2 , y2 ) and (1, –2) for (x1 , y1 ).
Use the slope formula.
The slope of the radius is . –3 4
Check It Out! Example 4 Continued
Because the slopes of perpendicular lines are negative reciprocals, the slope of the tangent is .
Holt Algebra 2
10-2 Circles
Use the point-slope formula.
Rewrite in slope-intercept form.
Substitute (5, –5 ) for (x1 , y1 ) and for m.
4 3
Check It Out! Example 4 Continued
Step 3. Find the slope-intercept equation of the tangent by using the point (5, –5) and the slope .
Holt Algebra 2
10-2 Circles
Check Graph the
circle and the line.
Check It Out! Example 4 Continued
The equation of the line that is tangent to 25 =
(x – 1)2 + (y + 2)2 at (5, –5) is .