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anonymous

  • one year ago

Write the general equation for the circle that passes through the points: (1, 7) (8, 6) (7, -1) You must include the appropriate sign (+ or -) in your answer. x^2 + y^2__x__y=0

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  1. mathstudent55
    • one year ago
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    The equation of a circle with center (x, y)and radius r is \((x - h)^2 + (y - k)^2 = r^2\) You have three given points. Substitute one point at a time into the equation above to get 3 three equations in three unknowns, h, k, and r. Solve the system of equations and substitute the values of h, k and r in the above equation.

  2. anonymous
    • one year ago
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    Ok so I have to put (1,7) into that equation you gave me??

  3. mathstudent55
    • one year ago
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    That is one way of doing it but it's a lot of work unless you have an electronic way of solving the system of equations.

  4. mathstudent55
    • one year ago
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    Here is another way of solving.

  5. mathstudent55
    • one year ago
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    Here is a circle with two points on it and a chord drawn through them. |dw:1435785536910:dw|

  6. mathstudent55
    • one year ago
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    |dw:1435785591709:dw|

  7. mathstudent55
    • one year ago
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    The perpendicular bisector of the chord passes through the center of the circle. |dw:1435785640579:dw|

  8. mathstudent55
    • one year ago
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    You are given three points on the circle. |dw:1435785694244:dw|

  9. mathstudent55
    • one year ago
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    Use a chord again and a perpendicular bisector again: |dw:1435785724236:dw|

  10. anonymous
    • one year ago
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    Ok so how can I figure out the equation by using that method?

  11. mathstudent55
    • one year ago
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    Where the perpendicular bisectors of the chords intersect is the center of the circle.

  12. mathstudent55
    • one year ago
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    Ok. Let's start. |dw:1435785943459:dw|

  13. mathstudent55
    • one year ago
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    |dw:1435786061026:dw|

  14. mathstudent55
    • one year ago
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    |dw:1435786091850:dw|

  15. mathstudent55
    • one year ago
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    Let's start with points (1, 7) and (8, 6) We need to find the equation of the perpendicular bisector of the segment with those endpoints.

  16. anonymous
    • one year ago
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    Ok

  17. mathstudent55
    • one year ago
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    1. Find the slope of the segment \(m = \dfrac{6 - 7}{8 - 1} = \dfrac{-1}{7} = -\dfrac{1}{7} \)

  18. mathstudent55
    • one year ago
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    The slope of the perpendicular bisector is m = 7

  19. anonymous
    • one year ago
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    ok

  20. mathstudent55
    • one year ago
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    The perpendicular bisector passes through the midpoint of the segment, so we need to find the midpoint of the segment.

  21. anonymous
    • one year ago
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    Ok

  22. mathstudent55
    • one year ago
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    midpoint = \((\dfrac{1 + 8}{2}, \dfrac{7 + 6}{2}) = (\dfrac{9}{2}, \dfrac{13}{2})\) |dw:1435786426317:dw|

  23. mathstudent55
    • one year ago
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    We have a point and the slope, so we can find the equation of that perpendicular bisector.

  24. mathstudent55
    • one year ago
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    \(y = mx + b\) \(\dfrac{13}{2} = 7(\dfrac{9}{2}) + b\) \(\dfrac{13}{2} = \dfrac{63}{2} + b\) \(b = -25\) \(y = 7x - 25\) This is the equation of the first perpendicular bisector of a chord.

  25. mathstudent55
    • one year ago
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    Now we need to choose another set of two points on the circle and do the same.

  26. anonymous
    • one year ago
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    Ok so how would I plug that in this equation? x^2 + y^2__x__y=0

  27. mathstudent55
    • one year ago
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    |dw:1435786849703:dw|

  28. mathstudent55
    • one year ago
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    |dw:1435786887498:dw|

  29. mathstudent55
    • one year ago
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    Now we have the chord in black in the last figure above. We need its slope and its midpoint.

  30. anonymous
    • one year ago
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    ok

  31. mathstudent55
    • one year ago
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    \(slope = m = \dfrac{-1 - 7}{7 - 1} = \dfrac{-8}{6} = -\dfrac{4}{3}\) The slope of the perpendicular bisector is \(\dfrac{3}{4} \)

  32. mathstudent55
    • one year ago
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    We need the coordinates of the midpoint of the chord. midpoint \(= (\dfrac{7 + 1}{2}, \dfrac{-1 + 7}{2}) = (4, 3)\)

  33. mathstudent55
    • one year ago
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    Since we now have the slope of the perpendicular bisector and a point it goes through, we can find the equation of the perpendicular bisector.

  34. anonymous
    • one year ago
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    ok

  35. mathstudent55
    • one year ago
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    \(y = mx + b\) \(3 = \dfrac{3}{4} \times 4 + b\) \(b = 0\) The eq of the perp bisector is: \(y = \dfrac{3}{4} x \)

  36. mathstudent55
    • one year ago
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    Now we have two equations of the two perpendicular bisectors of chords that we know intersect at the center pf the circle. We solve the system of equations: \(y = 7x - 25\) \(y = \dfrac{3}{4} x\) Since both equations are solved for y, we can equate the right sides and solve for x. \(7x - 25 = \dfrac{3}{4}x \) \(28x - 100 = 3x\) \(25x = 100\) \(x = 4\) \(y = \dfrac{3}{4} x = \dfrac{3}{4} \times 4 = 3\) Now we know the center is (4, 3).

  37. anonymous
    • one year ago
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    Ok

  38. mathstudent55
    • one year ago
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    |dw:1435787686770:dw|

  39. mathstudent55
    • one year ago
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    We can already fill in a little in the equation of the circle: The center is (h, k), and we know it is (4, 3). \((x - 4)^2 + (y - 3)^2 = r^2\)

  40. mathstudent55
    • one year ago
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    All we need now the to find the radius, so we can find r in the equation of the circle.

  41. anonymous
    • one year ago
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    Ok

  42. mathstudent55
    • one year ago
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    The radius of a circle is the distance between the center of the circle and any point on the circle. We know the center is (4, 3) Also, we were given three points on the circle. We now need to find the distance between the center and any of the three points we were given to find the radius.

  43. anonymous
    • one year ago
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    Ok

  44. mathstudent55
    • one year ago
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    Let's use point (8, 6) The distance between point (8, 6) on the circle, and point (4,3) the center is the radius of the circle. We use the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

  45. anonymous
    • one year ago
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    ok

  46. anonymous
    • one year ago
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    So it would be 5 right?

  47. mathstudent55
    • one year ago
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    \(d = \sqrt{(8 - 4)^2 + (6 - 3)^2} \) \(d = \sqrt{4^2 + 3^2} \) \(d = \sqrt{16 + 9} \) \(d = \sqrt{25} \) \(d = 5 \)

  48. mathstudent55
    • one year ago
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    Right. The radius is 5. Since the equation of a circle has r^2 on the right side, we can finish now: \((x - 4)^2 + (y - 3)^2 = 5^2\) or if you prefer \((x - 4)^2 + (y - 3)^2 = 25\)

  49. anonymous
    • one year ago
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    Ok

  50. mathstudent55
    • one year ago
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    |dw:1435788411933:dw|

  51. anonymous
    • one year ago
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    Thank you!

  52. mathstudent55
    • one year ago
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    You're welcome.

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