003 circle

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003 circle

  1. 1. LECTURE UNIT 003Conic Sections General Equation of Conics Ax2 + By2 + Dx + Ey + F = 0Circles Set of points which lie at a fixed distance ( ) from fixed point ( ). By definition of circle, we have CP = r or y P (x, y) (x - h)2 + (y - k)2 = r r c (h, k) Standard equation of a circle x (x - h)2 + (y - k)2 = r2 Where: r = radius (h, k) = center of the circle (x, y) = any point in the circle 1. Find the equation of the circle with given condition: with center at (-2, -3) and passing through (5, 1). 2. Find the equation of a circle of radius of 5 and center (1, 5). 3. Find the equation of the circle with ends of diameter at (-3, -5) and (5, 1). 4. Find the equation of the circle with center at the y-intercept of the line 4x + 3y - 12 = 0 and passing through the intersection of 3x + y = 6 and 3x - 2y = 10. 5. Show that the equation x2 - 2x + y2 + 6y = -6 represents a circle. Find its center and radius. 6. Find the center and radius whose equation is 4x2 + 4y2 - 8x + 4y + 1 = 0.Circle Tangent to the Line Ax + By + C = 0 y c (h, k) Ax + By + C = 0 r |Ah + Bk + C| x r= A2 + B2 7. Find the equation of the circle with center at (-1, 5) and tangent to 3x + y - 6 = 0. “One of the marks of true greatness is the ability to develop greatness in others.”
  2. 2. The General Equation of a Circle The equation x2 + y2 + Dx + Ey + F = 0 is the general equation of the circle, where D, E and F are the arbitrary constants. To sketch the graph, reduce the equation to its standard form. 2 2 D E D2 + E2 (x + Dx + 2 ) + (y + Ey + 2 ) = -F + 4 4 2 2 2 2 ( x+ D ) ( + y+ E ) = D + E - 4F 2 2 4 D2 + E2 - 4F r2 = 4 Note: r2 < 0 no graph (imaginary circle) r2 = 0 single point r2 > 0 graph is a circle Sketch the graph: 8. x2 + y2 - 6x + 2y + 1 = 0 9. x2 + y2 - 5x + 3y + 20 = 0 10. 4x2 + 4y2 - 8x + y - 4 = 0 11. x2 + y2 + 10x + 25 = 0 12. x2 + y2 - 8x + 6y - 11 = 0 13. 2x2 + 2y2 - 7x + y + 100 = 0 14. 3x2 + 3y2 - 9x + 3y - 24 = 0 15. x2 + y2 - 2x + 4y + 5 = 0Circle Determined by Three Conditions x2 + y2 + Dx + Ey + F = 0 (General equation of a circle) Where: D, E and F are the arbitrary constants Use this equation when there are 3 points in the circle (x - h)2 + (y - k)2 = r2 (Standard equation of a circle) Where: h, k and r are the arbitrary constants Use this equation when the given radius of the circle is tangent to a line. 16. Find the equation of the circle passing through (1, -2), (3, 1) and (2, -1). 17. Passing through the origin and the points (5, -2) and (3, 3). Find the equation of the circle. 18. With center on the y=axis and passing through (2, -3) and (-1, 4). “Great leaders are never satisfied with current levels of performance. They are relentlessly driven by possibilities and potential achievements.”
  3. 3. Family of Circles 19. Find the equation of the family of circles with center at (3, -2). 20. Find the equation of the family of circles with center on the x-axis.To find the the equation of the family of circles passing through the intersection of 2 circles whose equation are: x2 + y2 + D1x + E1y + F1 = 0 (C1) x2 + y2 + D2x + E2y + F2 = 0 (C2) C1 and C2 are non concentric circlesEquation of the family is x2 + y2 + D1x + E1y + F1 + k(x2 + y2 + D2x + E2y + F2) = 0 Where: k = 0 Will result to another original equation. k = -1 Will result to a linear equation. If k = -1 the resulting equation is the equation of the radical axis (R.A.) of 2 circles.Radical axis - a line perpendicular to the line joining the centers of the 2 circles. y y R.A. R.A. C1 (h1, k1) C1 x x C2 (h2, k2) C2 21. Find the equation of the circle that passes through the intersection of x2 + y2 - 2x + 4y - 4 = 0 and x2 + y2 + 4x + 6y - 3 = 0. (a) Find the equation of one member (b) find the equation of the radical axis and (c) find the equation of the member that passes through (-3, 4). 22. Find the equation of the circle that passes through the intersection of x2 + y2 + 3x - y - 5 = 0 and x2 + y2 - 2x + 4y - 11 = 0 and the point (-1, 2). “Enthusiasm is contagious. It’s difficult to remain neutral or indifferent in the presence of a positive thinker.”
  4. 4. CIRCLES Example 1: Find the equation of the circle with given condition: with center at (-2, -3) and passing through (5, 1). Solution: From; (x - h)2 + (y - k)2 = r2 (5 + 2)2 + (1 + 3)2 = r2 r2 = 65 (5, 1) r = 65 r = 8.06226 Hence; (-2, -3) (x + 2)2 + (y + 3)2 = 65 r = 8.06226 (x + 2)2 + (y + 3)2 = 65 Example 5: Show that the equation x2 - 2x + y2 + 6y = -6 represents a circle. Find its center and radius. Solution: x2 - 2x + y2 + 6y = -6 Completing squares: (x2 - 2x + 1) + (y2+ 6y + 9) = -6 + 1 + 9 (x - 1)2 + (y + 3)2 = 4 C (1, -3) r=2 Example 7: Find the equation of the circle with center at (-1, 5) and tangent to 3x + y - 6 = 0. Solution: (x + 1)2 + (y - 5)2 = r2 (0, 6) (-1, 5) | 3(-1) + 1(5) - 6| r= 3x + y - 6 = 0 32 + 12 4 r =- 10 (2, 0) r2 = 8 5 (x + 1)2 + (y - 5)2 = 8 5 “If you stop learning today, you stop leading tomorrow.”
  5. 5. Example 8: Sketch the graph x2 + y2 - 6x + 2y + 1 = 0Solution: (x2 - 6x + 9) + (y2 + 2y + 1) = -1 + 1 + 9 (x - 3)2 + (y + 1)2 = 9 C (3, -1) (3, -1) r=3 r=3Example 16: Find the equation of the circle passing through (1, -2), (3, 1) and (2, -1).Solution: P1 (1, -2) D - 2E + F = -5 eq. 1 P2 (3, 1) 3D + E + F = -10 eq. 2 P3 (2, -1) 2D - E + F = -5 eq. 3 By elimination and substitution, we find that: D=5 E = -5 (2.5, 2.5) F = -20 Therefore: r = 5.7 (3, 1) x2 + y2 + 5x - 5y - 20 = 0 (2, -1) (1, -2) (x + 2.5)2 + (y - 2.5)2 = 65/2Example 16: With center on the y=axis and passing through (2, -3) and (-1, 4).Solution: Center on the y-axis, C (0, k) x2 + (y - k)2 = r2 eq. 1 (-1, 4) (2, -3) 4 + (-3 - k)2 = r2 13 + 6k + k2 = r2 eq. 2 (-1, 4) 1 + (4 - k)2 = r2 17 - 8k + k2 = r2 eq. 3 (0, 2/7) Solving simultaneously; k = 2/7 r2 = 725/49 (2, -3) Required equation: X2 + ( y - 2 ) = 725 2 7 49 "Anger and haste hinder good counsel.”
  6. 6. Example 22: Find the equation of the circle that passes through the intersection of x2 + y2 + 3x - y - 5 = 0 and x2 + y2 - 2x + 4y - 11 = 0 and the point (-1, 2).Solution: x2 + y2 + 3x - y - 5 + k (x2 + y2 - 2x + 4y - 11) = 0 equation of the family at (-1, 2) we find k = 5/4 Substitute the value of k, and determine the equation of the circle: X2 + y2 + 3x - y - 5 + (5/4)(x2 + y2 - 2x + 4y - 11) = 0 Solving simultaneously, the desired equation of the circle is: 9x2 + 9y2 + 2x + 16y - 75 = 0 To determine the equation of the radical-axis, k = -1 5x - 5y + 6 = 0 x2 + y2 + 3x - y - 5 = 0 (-1, 2) 5x - 5y + 6 = 0 (-1.5, 0.5) (-1/9, -8/9) (1, -2) 9x2 + 9y2 + 2x + 16y - 75 = 0 x2 + y2 - 2x + 4y - 11 = 0Example 19: Find the equation of the family of circles with center at (3, -2).Solution: (x - 3)2 + (y + 2)2 = r2 equation of the family r=1 (x - 3)2 + (y + 2)2 = 1 equation of first member r=3 (x - 3)2 + (y + 2)2 = 9 equation of the second memberExample 20: Find the equation of the family of circles with center on the x-axis.Solution: (x - h)2 + y2 = r2 equation of the family r = 4, h = -2 (x + 2)2 + y2 = 16 equation of one member “Alone we can do so little; together we can do so much.”

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