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23.2 Equations of Circles

23. Locus. Case Study. 23.1 Concept of Loci. 23.2 Equations of Circles. 23.3 Intersection of a Straight Line and a Circle. Chapter Summary. The satellite should move in a way that prevents itself from deviating from the path. Do you know how the Chang’e 1 Satellite orbited the Moon?.

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23.2 Equations of Circles

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  1. 23 Locus Case Study 23.1Concept of Loci 23.2 Equations of Circles 23.3 Intersection of a Straight Line and a Circle Chapter Summary

  2. The satellite should move in a way that prevents itself from deviating from the path. Do you know how the Chang’e 1 Satellite orbited the Moon? Case Study On 24 October 2007, China’s first lunar orbited ‘Chang’e 1 Satellite’ was launched from Xichuan. The satellite orbited the Moon at an altitude of 200 km above the lunar surface. It carried out a one-year lunar exploration mission.

  3. 23.1 Concept of Loci A. Description of Loci If a point is moving under a specific condition, then its path is called the locus of the moving point. Consider a football rolling on a horizontal and smooth ground, the locus of the centre of the football is a straight line as shown in the figure below. When a pendulum swings to and fro, the locus of its tip is an arc of a circle with the length of the pendulum as the radius. In mathematics, locus is the set of points that satisfy or are determined by some specific conditions. It can be a straight line, curve, polygon or circle.

  4. 23.1 Concept of Loci A. Description of Loci Case 1 : A point R moves in a plane such that it always maintains a fixed distance of 3 cm from a fixed line L. Locus : The locus of the point R is a pair of straight lines which are (1) parallel to L; (2) on one of the sides of L respectively; (3) equidistant from L with a distance of 3 cm. Case 2 : A point Q moves in a plane such that it always maintains an equal distance of 2 cm from two parallel lines AB and CD. Locus : The locus of the point Q is a straight line which is (1) parallel to both AB and CD; (2) in the middle of AB and CD; (3) equidistant from AB and CD with a distance of 2 cm.

  5. P  O  23.1 Concept of Loci A. Description of Loci Example 23.1T Describe the locus of the moving point P under each of the following conditions. (a)P is always 3 cm from the origin O. (b)AB is a straight line and ÐPAB  ÐPBA. Solution: (a) A circle centred at O with radius 3 cm. (b) The perpendicular bisector of AB.

  6. 23.1 Concept of Loci A. Description of Loci Example 23.2T In the figure, l1 and l2 are two parallel lines 6 cm apart. A moving point P is equidistant from l1 and l2. Sketch the locus of P. Solution: The locus is a line which is parallel tol1 and l2, and is 3 cm froml1 andl2.

  7. 23.1 Concept of Loci A. Description of Loci Example 23.3T In the figure, AB is a straight line. The locus of a point P is formed by the centres of all the circles with radius 1 cm and touch AB. Describe the locus of P. Solution: A pair of lines parallel to AB and are 1 cm away from it.

  8. 23.1 Concept of Loci A. Description of Loci Example 23.4T AB is a line segment. A moving point P passes through A and B and moves in a way such that ÐAPB  130°. Describe and sketch the locus of P. Solution: The locus of P is a minor arc APB of a circle.

  9. 23.1 Concept of Loci B. Describe Loci in Algebraic Equations Besides verbal description of the locus, we can also describe the locus of a point in an algebraic equation.

  10. Since the distance between P and Q is , 23.1 Concept of Loci B. Describe Loci in Algebraic Equations Example 23.5T A moving point P is always units from . Express the locus of P in algebraic form. Solution: Let (x, y) be the coordinates of P. The locus of P is 4x2 + 4y2 – 4x + 8y – 3  0.

  11. 23.1 Concept of Loci B. Describe Loci in Algebraic Equations Example 23.6T Consider a point K(5, 0) and a horizontal line y  –2. P is a moving point such that its distance from the line is equal to its distance from the point K. Express the locus of P in the form y  ax2 + bx + c. Solution: Let (x, y) be the coordinates of P. Since the distance from P to the line is equal to the distance from P to the point (x, 2) on the line, distance from P to (x, 2)  distance from P to K(5, 0) The locus of P is .

  12. 23.2Equations of Circles A. Circles In section 23.1, we learnt that the locus of a moving point that keeps a fixed distance from a fixed point is a circle. Note: The fixed distance is the radius and the fixed point is the centre of the circle. In this section, we put the circle on the coordinate plane and hence find the equation of the circle.

  13. If the centre is the origin (0, 0), then the equation of the circle is x2 + y2  r2. 23.2Equations of Circles A. Circles Suppose the centre is C(h, k) and the radius is r. Let P(x, y) be the moving point on the circle, then from the distance formula, we have Taking square on both sides, we have (x  h)2 + (y  k)2  r2. Thus, the equation of the circle is (x  h)2 + (y  k)2  r2, where centre  (h, k) and radius  r. Note: This equation is called the centre-radius form of the circle.

  14. 23.2Equations of Circles B. General Form of Equations of Circles If we expand the equation of the circle (x  h)2 + (y  k)2  r2, the equation can be expressed in the form x2  2hx + h2 + y2  2ky + k2  r2 x2 + y2  2hx  2ky + (h2 + k2  r2)  0 i.e., x2 + y2 + Dx + Ey + F 0, where D 2h, E 2k and Fh2 + k2  r2. This is called the general form of the equation of a circle. Notes: 1. The coefficients of x2 and y2 are both equal to 1. 2. D, E and F can be any real numbers. 3. The right hand side of the general form of the equation of a circle is 0.

  15. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.7T If a circle is centred at (4, 2) and it passes through (2, 2), find the equation of the circle. Express the answer in the general form. Solution: The radius of the circle 2 (4)  2. The equation of the circle:

  16. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.8T Given that A(3, 5) and B(9, 7) are the two end points of a diameter of a circle. (a)Find the equation of the circle in general form. (b)Find the coordinates of the points of intersection of the circle and the x-axis. Solution: (a) Centre (b) When y  0, we have x  4 or 2 Radius  The points of intersection of the circle and the x-axis are (4, 0) and (2, 0). The equation of the circle:

  17. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.9T A circle passes through three points O(0, 0), A(6, 0) and B(0, 10). (a)Find the equation of the circle in the general form. (b)Does the point I(1, 2) lie on the circle? Solution: (a) Let the equation of the circle be x2 + y2 + Dx + Ey + F 0 …(*) Since the three points O, A and B must satisfy (*),

  18. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.9T A circle passes through three points O(0, 0), A(6, 0) and B(0, 10). (a)Find the equation of the circle in the general form. (b)Does the point I(1, 2) lie on the circle? Solution: Substituting (1) into (2) and (3), From (4), 6D36 D6 From (5), 10E 100 E 10  The equation of the circle is x2 + y2 – 6x + 10y 0.

  19. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.9T A circle passes through three points O(0, 0), A(6, 0) and B(0, 10). (a)Find the equation of the circle in the general form. (b)Does the point I(1, 2) lie on the circle? Solution: (b) Substituting (1, 2) into the equation x2 + y2 – 6x + 10y 0. L.H.S.  (1)2 + (2)2  6(1) + 10(2) 9  R.H.S. ∵ (1, 2) does not satisfy the equation of the circle.  I(1, 2) does not lie on the circle.

  20. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.10T The centre of a circle lies on the straight line x + 2y + 1  0. The circle passes through S(1, 3) and T(2, 0).   (a) Find the equation of the circle in the general form.   (b) If PS is a diameter of the circle, find the coordinates of P. Solution: (a) Let R(h, k) be the centre of the circle. Since the centre lies on x + 2y + 1  0, we have h + 2k + 1  0…………(1)

  21. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.10T The centre of a circle lies on the straight line x + 2y + 1  0. The circle passes through S(1, 3) and T(2, 0).   (a) Find the equation of the circle in the general form.   (b) If PS is a diameter of the circle, find the coordinates of P. Solution: (1)  (2): Substituting k  2 into (2), we have The equation of the circle:  The centre  (3, 2) Radius

  22. 23.2Equations of Circles B. General Form of Equations of Circles Example 23.10T The centre of a circle lies on the straight line x + 2y + 1  0. The circle passes through S(1, 3) and T(2, 0).   (a) Find the equation of the circle in the general form.   (b) If PS is a diameter of the circle, find the coordinates of P. Solution: (b) Let (x, y) be the coordinates of P. Since (3, 2) is the mid-point of P and S, we have The coordinates of P are (5, 7).

  23. For the equation of a circle x2 + y2 + Dx + Ey + F  0, we have • centre (b) radius 1.If > 0, then the radius is a real number. This kind of circle is known as a real circle. 2.If  0, then the equation represents a circle with zero radius. 3.If < 0, then the radius is not a real number. This kind of circle is known as a point circle. This kind of circle is known as an imaginary circle. 23.2Equations of Circles C. Features of Equations of Circles Remarks:

  24. 23.2Equations of Circles C. Features of Equations of Circles Example 23.11T Consider a circle: x2 + y2 + 5x – 10y + 15  0 Determine whether the point A(1, –1) lies on, inside or outside the circle. Solution: Centre Radius Distance between the point A and the centre  radius A(1, 1) lies outside the circle.

  25. 23.2Equations of Circles C. Features of Equations of Circles Example 23.12T The following are the equations of two circles: C1 : x2 + y2 – 2x – 14y + 46  0 C2: x2 + y2 – 12x + 10y – 164  0 (a) Find the centres and radii of the two circles. (b) Hence show that the two circles touch each other internally. Solution: (a) For C1, centre and radius For C2, centre and radius

  26. 23.2Equations of Circles C. Features of Equations of Circles Example 23.12T The following are the equations of two circles: C1 : x2 + y2 – 2x – 14y + 46  0 C2: x2 + y2 – 12x + 10y – 164  0 (a) Find the centres and radii of the two circles. (b) Hence show that the two circles touch each other internally. Solution: (b) Distance between the centres Difference of the radii of the two circles  15 – 2  13  Distance between the centres  The two circles touch each other internally.

  27. 23.3Intersection of a Straight Line and a Circle In the same coordinate plane, there are three cases showing the relationship between the graphs of a circle x2 + y2 + Dx + Ey + F  0 and a straight line y  mx + c. Case 1 : intersect at two distinct points; Case 2 : intersect at one point only; Case 3 : no point of intersection. Notes: For case 2, the straight line is called a tangent to the circle.

  28. 23.3Intersection of a Straight Line and a Circle Without the actual drawing of the graphs, the number of points of intersection of the two graphs can be determined algebraically by carrying out the following steps. Step 1:Use the method of substitution to eliminate one of the unknowns (either x or y) of the simultaneous equations. We can then obtain a quadratic equation in one unknown. Substituting (1) into (2), x2 + (mx + c)2 + Dx + E(mx + c) + F 0 x2 + m2x2 + 2mcx + c2 + Dx + Emx + Ec + F 0 (1 + m2 )x2 + (2mc + D + Em)x + (c2 + Ec + F) 0 ... (*) • Step 2:Evaluate the discriminant (D) of the quadratic equation (*). • If D > 0, there are two points of intersection. • If D  0, there is only one point of intersection. • If D < 0, there is no point of intersection.

  29. 23.3Intersection of a Straight Line and a Circle Example 23.13T Find the coordinates of the points of intersection of the straight line 2x – y – 1  0 and the circle x2 + y2 – 3x + y – 10  0. Solution: Consider the simultaneous equations of the straight line and the circle. When x  1, y  2(1)  1  3 Substituting (1) into (2), we have When x  2, y  2(2)  1  3 The points of intersection are (1, 3) and (2, 3). x  1 or 2

  30. 23.3Intersection of a Straight Line and a Circle Example 23.14T If the straight line y – 3x – 1  0 meets the circle x2 + y2 + 2x + 4y + k  0 at two distinct points, find the range of values of k. Solution: Consider the simultaneous equations of the straight line and the circle. Substituting (1) into (2), Since the straight line meets the circle at two distinct points, the discriminant of (*) is greater than 0.  The range is k < 5.

  31. 23.3Intersection of a Straight Line and a Circle Example 23.15T Given a circle (x – 1)2 + (y + 2)2  k. If P(4, –3) lies on the circle, find (a)the centre of the circle, (b)the value of k, (c)the equation of the tangent to the circle at P. Solution: (a) Centre (b) Substituting (4, 3) into the equation

  32. 23.3Intersection of a Straight Line and a Circle Example 23.15T Given a circle (x – 1)2 + (y + 2)2  k. If P(4, –3) lies on the circle, find (a)the centre of the circle, (b)the value of k, (c)the equation of the tangent to the circle at P. Solution: (c) Slope of the line joining the centre and P Slope of the tangent  Equation of the tangent:

  33. Chapter Summary 23.1 Concept of Loci 1. A locus is the set of all points that satisfy the given specified conditions. 2. The locus of points can be expressed in algebraic equations.

  34. 3. From the general form, we have (a) centre  and (b) radius  Chapter Summary 23.2 Equations of Circles 1. A circle can be expressed by the centre-radius form (x  h)2 + (y  k)2  r2, where centre  (h, k) and radius  r. 2. The general form of a circle is x2 + y2 + Dx + Ey + F 0. 4. If the distance between the centre of a circle and a point P is (a) smaller than the radius of the circle, then P lies inside the circle; (b) equal to the radius of the circle, then P lies on the circle; (c) greater than the radius of the circle, then P lies outside the circle.

  35. Chapter Summary 23.3 Intersection of a Straight Line and a Circle 1. There are three cases showing the relationship between a straight line and a circle: (a) Intersect at two points (b) Intersect at one point (c) No point of intersection We can use the discriminant to determine the number of the points of intersection. 2. The coordinates of the point(s) of intersection can be found by solving the simultaneous equations of a straight line and a circle. 3. If a straight line touches the circle at one point only, then the straight line is a tangent to the circle.

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