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- 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. 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. 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. 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. 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. 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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