Analytic Geometry Section 2-6: Circles

Transcription

1 Analytic Geometry Section 2-6: Circles Objective: To find equations of circles and to find the coordinates of any points where circles and lines meet. Page 81

2 Definition of a Circle A circle is the set of all points in a plane equidistant from a fixed point called the center point. We can derive the equation directly from the distance formula. If we place the center point on the origin point, the equation of a circle with center point (0, 0) and radius r is: x 2 + y 2 = r 2

3 Circles and Points of Intersection circle: in a plane, the set of points equidistant from a given point, called the center. radius: any segment whose endpoints are the center and a point on the circle. If the circle is centered at (0, 0), and the radius is r, then the distance to any point, (x, y) on the circle (using the distance formula) is x 0 2 y 0 2 r square both sides... (x,y) x 0 2 y 0 2 r 2 now simplify... x 2 y 2 r 2 This is the standard form of a circle with center (0,0) and radius r.

4 Example: Write an equation of the circle with its center at the origin with the point ( 3, 4) on the circle. Use the Standard Equation of the Circle to find the radius of the circle... x 2 y 2 r r r 2 25 r 2 x 2 y 2 25

5 (x - h) 2 + (y - k) 2 = r 2 With center at (h, k) and radius r. Here are some examples:

6 Sample Problems 1) Find the center, radius and graph the equation: (x - 2) 2 + (y + 5) 2 = 17 Solution: Center point : (2, -5), radius =

7 Find the center and radius of the circle with the following equation: 100x y 2 100x + 240y 56 = 0. This is the given equation. Move the loose number over to the other side. Group the x-stuff and y-stuff together. Divide off by whatever is multiplied on the squared terms.

8 Find the center and radius of the circle with the following equation: 100x y 2 100x + 240y 56 = 0. Take the coefficient on the x-term, multiply by one-half, square, and add inside the x-stuff and also to the other side. Do the same with the y-term. Convert the left-hand side to squared form, and simplify the right-hand side. If necessary, fiddle with signs and exponents to make your equation match the circle equation's format.

9 Find the center and radius of the circle with the following equation: 100x y 2 100x + 240y 56 = 0. Read off the answer. The center is at ( 1 / 2, 6 / 5 ) and the radius is 3 / 2.

10 2) x 2 + y 2-8x + 4y - 8 = 0 Find center, radius and graph. Solution: We need to put the equation into the correct form. We will do this by completing the square!! (x 2-8x ) + (y 2 + 4y ) = 8 Complete the square! See 1.6 (x 2-8x + 16) + (y 2 + 4y + 4) = (x - 4) 2 + (y + 2) 2 = 28 Now it's in the correct form!! Center point (4, -2) with radius =

11 4) Sketch the graph of Solution: The above graph is part of a circle. Why?

12 3) Find the intersection of the line y = x - 1 and the circle x 2 + y 2 = 25. Solution: x 2 + (x - 1) 2 = 25 x 2 + x 2-2x + 1 = 25 2x 2-2x - 24 = 0 x 2 - x - 12 = 0 Now factor (x - 4)(x + 3) = 0 x = 4 or x = -3 Substituting back in to find y gives the following points: (4, 3) and (-3, -4)

13 Example: Find the points of intersection, if any, of the graphs of x 2 + y 2 = 25 and y = 2/3 x + 2 Use substitution x 2 + (2/3 x + 2) 2 = 25 simplify... x 2 + 4/9 x 2 + 8/3 x + 4 = 25 13/9 x 2 + 8/3 x + 4 = 25 13/9 x 2 + 8/3 x 21 = 0 multiply by 9 13x x 189 = 0 factor (13x + 63)(x 3) = x + 63 = 0 or x 3 = 0 x = 63 or x = 3 13 Now, find the y-coordinate for each x-coordinate... y = 2/3( 63/13) + 2 y = 2/3 (3) + 2 y = 126/ y = y = 126/ /39 y = 4 y = 48/39 y = 16/13 The points of intersection are ( 63/13, 16/13) and (3, 4)

14 To find the Intersection of a Line and a Circle Algebraically: 1. Solve the linear equation for y in terms of x (or x in terms of y). 2. Substitute this expression for y (or x) in the equation of the circle. Then solve the resulting quadratic equation. 3. Substitute each real x-solution from step 2 in the linear equation to get the corresponding value of y (or vice versa). Each point (x,y) is an intersection point. 4. You can check your result by substituting the coordinates of the intersection points in the two original equations.

15 Example: Find the equation of the circle that passes through the points P(1, -2), Q(5, 4), and R(10, 5). 2 2 x y Dx Ey F Q R Using Points P 1,-2, 5,4,and 10,5 1 4 D 2E F D 4E F D 5E F 0 D 18, E 6, F 25. x y 18x 6y 25 0

16 Example A circle is tangent to the line 2x y + 1 = 0 at the point (2, 5), and the center is on the line x + y= 9. Find the equation of the circle. 8 6 (2,5) B 4 (6,3)

17 The line through (2, 5) and perpendicular to the line 2x y + 1 = 0 passes through the center of the circle. The equation of this line is x + 2y = 12. Hence, the solution of the system X + 2y = 12 X + y = 9 Yields the coordinates of the center. Accordingly, the center is at (6, 3). The distance from this point to (2, 5) is 20. The equation of the circle, therefore, is (x 6) 2 + (y 3) 2 = 20.

18 Example A triangle has its sides on the lines x + 2y 5 = 0, 2x y 10 = 0, and 2x + y + 2 = 0. Find the equation of the circle inscribed in the triangle d1 p d3-2 d

19 2 x ' y ' 2 2 x ' y ' ( 1) x ' y ' 2 2 x ' y ' 10 4 x ' x ' 8 x ' Bisector 2 x ' y ' 10 x ' 2 y ' x ' y ' 10 x ' 2 y ' 5 x' 3 y' 5 0 x' 3 y' 5 Bisector

20 x ' 2 x'-3 y' y ' 5-3 y ' 3 y ' -1 Intersection point of the bisectors is (2, -1) This is the center of the circle. The distance from the center to the side of the circle is x 2 y 1 5.

21 Example: Find the equation of a circle that passes through the points (2,-1), (0,2), and (1,1).

22 Example: Find the equation of the circle passing through the points (1,-2) and (5,3) and having its center on the line x y + 2 =0.

23 Example: A circle is tangent to the line 2x y + 1 = 0 at the point (2,5), and the center of the circle falls on the line x + y = 9. What is the equation of the circle?

24 Example: Find the equation of the circle inscribed in the triangle determined by the lines, L1: 2x 3y + 21=0, L2: 3x 2y 6 = 0, L3: 2x + 3y + 9 = 0.

25 Example: Derive the equation of the circle circumscribing the triangle determined by the lines, x + y = 8, 2x + y = 14, 3x + y = 22.

26 Example: Derive the equation of the circle which passes through the point (-2, 1) and is tangent to the line 3x 2y 6 = 0 at the point (4,3).

27 Homework Assignment Page Problems 1 49 odd