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Math Help - Co-ordiante geometry

  1. #1
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    Co-ordiante geometry

    Can anyone help me with these with step by step answers.
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    Hello, novadragon849!

    Here are a few of them . . .

    You're expected to know the general equation of the circle.
    . . The circle with center C(h,k) and radius r has the equation:
    . . . . (x - h) + (y -k) .= .r


    1) Find the coordinates of the center and the radius of the circle:
    . . . . . (x - 7) + (y + 2) .= .64
    Do you know how to use that general equation?

    The center is (7,-2) and the radius is 8.



    2) The points P(10,-4) and Q(2,2) are the ends of a diameter.
    . . Find the equation of the circle.
    The center is the midpoint of diameter PQ.

    We have: .x = (10 + 2) = 6 .and .y = (-4 + 2) = -1
    . . Hence, the center is: C(6,-1)

    The radius is the distance CP or CQ.
    . . CQ .= .√[(2 - 6) + (2 + 1)] .= .√[16 + 9] .= .√[25] .= .5

    Therefore, the equation is: .(x - 6) + (y + 1) .= .25



    3) The point (-2,3) lies on the circle: .(x - 4) + (y + 1) .= .r
    . . Find the exact value of r.

    If (-2,3) lies on the circle, then x = -2, y = 3 satisfies the equation.

    We have: .(-2 -4) + (3 + 1) .= .r . . r .= .(-6) + 4 .= .36 + 16 .= .52

    . . . . . . - . - . . . .__ . . . . .__
    Therefore: .r .= .√52 .= .2√13

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  3. #3
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    Thanks but has anyone got the answer to the rest of the questions.
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    Quote Originally Posted by novadragon849 View Post
    Can anyone help me with these with step by step answers.
    4. (a) p(-1,2), C(4,5)

    Equation of a circle is of the form:

    (x - h)^2 + (y - k)^2 = r^2
    where (h,k) is the center of the circle and r is the radius
    using (x,y) = (-1,2) and (h,k) = (4,5)
    => (-1 - 4)^2 + (2 - 5)^2 = r^2
    => 25 + 9 = r^2
    => 34 = r^2
    => r = sqrt(34)

    thus the equation of the circle is:
    (x - 4)^2 + (y - 5)^2 = 34


    (b) (x - 4)^2 + (y - 5)^2 = 34
    => 2(x - 4) + 2(y - 5) y' = 0
    => y' = -(x - 4)/(y - 5)

    at P(-1,2), y' = -(-1 -4)/(2 - 5)
    = y' = -5/3 ......gradient of the tangent at P

    (c) Using y - y1 = m(x - x1)
    => y - 2 = -(5/3)(x + 1)
    => y - 2 = -(5/3)x - 5/3
    => (5/3)x + y - 1/3 = 0 ...........equation of the tangent line at P


    5. P(2,2), Q(1,1), R(7,-3)

    the perpendicular bisector of a line connecting two points is a line passing throught the midpoint of the line connecting two points perpendicular to the line.

    (a) slope of PQ = (1 - 2)/(1 - 2) = 1
    => slope of perpendicular bisector = -1

    midpoint(PQ) = ((x2 + x1)/2, (y2 + y1)/2) = ( 3/2 , 3/2)

    so equation of perpendicular bisector of PQ:

    y - y1 = m(x - x1)
    => y - 3/2 = -1(x - 3/2)
    => y - 3/2 = -x + 3/2
    => y = -x + 3

    slope of PR = (-3 -2)/(7 - 2) = -1
    => slope of perpendicular bisector = 1

    midpoint(PR) = ((7 + 2)/2 , (-3 + 2)/2) = ( 9/2 , -1/2 )

    equation of bisector:
    y - y1 = m(x - x1)
    => y + 1/2 = x - 9/2
    => y = x - 5

    b) The center (I believe) will be where these lines intersect:

    so we want:

    x - 5 = -x + 3
    => 2x = 8
    => x = 4

    when x = 4

    y = 4 - 5 = -1

    so center of circle is -1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by novadragon849 View Post
    Can anyone help me with these with step by step answers.
    6. Note that C is on the x-axis and D is on the y-axis

    Using (x + 3)^2 + (y - 3)^2 = 90

    for x-intercept, y = 0
    => (x + 3)^2 + (0 - 3)^2 = 90
    => x^2 + 6x + 9 + 9 - 90 = 0
    => x^2 + 6x - 72 = 0
    => (x + 12)(x - 6) = 0
    => x = -12 or x = 6
    since we are concerned with the positive region of the x-axis, x = 6
    so C(6,0) so c = 6

    for y-intercept, x = 0
    => (0 + 3)^2 + (y - 3)^2 = 90
    => 9 + y^2 - 6y + 9 - 90 = 0
    => y^2 - 6y - 72 = 0
    so y = 12 or y = -6
    we want the positive region of the y-axis, so y = 12
    D(0,12) so d = 12

    7. x^2 + y^2 = 80
    => y = sqrt(80 - x^2) we show that the line y = 30 - 2x does not meet this circle.

    set the y's equal, we get:

    30 - 2x = sqrt(80 - x^2)
    => (30 - 2x)^2 = 80 - x^2
    => 900 - 120x + 4x^2 = 80 - x^2
    => 5x^2 - 120x + 820 = 0
    By the quadratic formula:
    x = [120 +/- sqrt(14400 - 16400)]/10
    => x = [120 +/- sqrt(-2000)]/10

    I don't feel like simplifying any further, i'm tired. just note that there is a negative number under the square root, so the two graphs do not meet at any real number coordinates and therefore do not meet.
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  6. #6
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    Hello, novadragon849!

    I hope you're learning something from my explanations.
    . . I really hate doing someone's homework.


    4) The point P(-1,2) lies on the circle with center C(4,5).
    (a) Find the equation of the circle.
    We already have: .(x - 4) + (y - 5) .= .r

    All we need is the radius.
    It's the distance between P and C.
    I'll let you find it . . .


    (b) Find the gradient of the tangent at P.
    This is a bit more subtle.

    The tangent to a circle is perpendicular to the radius at that point.

    The radius is CP, and its slope is: .(5 - 2)/(4 + 1) .= .3/5

    Therefore, the slope (gradient) of the tangent is: .-5/3


    (c) Find the equation of the tangent at P in the form: ax + by + c = 0
    We know point P(-1,2) and the slope, m = -5/3

    Use the Point-Slope Formula to write the equation of that line:
    . . y - 2 .= .(-5/3)(x + 1) . . 5x + 3y - 1 .= .0




    5) The points P(2,2), Q(1,1) and R(7,-3) lies on a circle.
    (a) Find the equation of the perpendicular bisector of: (i) PQ .(ii) PR
    (i) The midpoint of PQ is: .x = (2 + 1) = 3/2, . y = (2 + 1) = 3/2
    The slope of PQ is: .(2 - 1)/(2 - 1) .= .+1
    The perpendicular slope is: .m = -1
    . . The equation is: .y - 3/2 .= .-1(x - 3/2) . . y .= .-x + 3

    (ii) The midpoint of PR is: .x = (2 + 7) = 9/2, . y = (2 - 3) = -1/2

    The slope of PR is: .(-3 -2)/(7 - 2) .= .-1
    The perpendicular slope is: .m = +1
    The equation is: .y + 1/2 .= .1(x - 9/2) . . y .= .x - 5


    (b) Hence, find the coordinates of the center of the circle.
    The center is at the intersection of the two perpendicular bisectors.

    Solve the system of equations: .y .= .-x + 3, . y .= .x - 5

    I'll let you do that . . .




    6) The circle (x + 3) + (y - 3) .= .90
    meets the positive coordinates axes at C(c,0) and d(0,d).

    (a) Find the values of C and D.
    For an x-intercept, let y = 0.
    . . (x + 3) + (0 - 3) .= .90 . . (x + 3) + 9 .= .90
    . . (x + 3) .= .81 . . x + 3 .= .9 . . x .= .6, -12
    Therefore: .C(6,0)

    For a y-intecept, let x = 0.
    . . (0 + 3) + (y - 3) .= .90 . . 9 + (y - 3) .= .90
    . . (y - 3) .= .81 . . y - 3 .= .9 . . y .= .12, -6
    Therefore: .D(0,12)


    (b) Find the area of ∆OCD, where O is the origin.
    ∆OCD is a right triangle with base 6 and height 12.

    You finish it . . .

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