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    just wondering if anyone can help me....

    1) FIND THE EQUATIONS TO THE TANGENTS TO THE CIRCLE

    x^2 + y^2 = 25

    FROM THE POINT P(2,11), OUTSIDE THE CIRCLE


    and...


    2) SHOW THAT THE STRAIGHT LINE

    x - 3y - 10 = 0

    IS A TANGENT TO THE CIRCLE

    x^2 + y^2 = 10

    BY: (I) FINDING THE DISTANCE OF THE LINE FROM THE ORIGIN
    (II) SHOWING THAT THE LINE MEETS THE CIRCLE AT ONE POINT ONLY

    FURTHER, DETERMINE THE EQUATION TO THE NORMAL TO THE CIRCLE THROUGH THE POINT OF CONTACT OF THE TANGENT






    got really stuck...thanks for any help
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    ooh and particularly the first one...im really stuck
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    (Original post by CCDB)
    just wondering if anyone can help me....

    1) FIND THE EQUATIONS TO THE TANGENTS TO THE CIRCLE

    x^2 + y^2 = 25

    FROM THE POINT P(2,11), OUTSIDE THE CIRCLE
    you should know \frac{dy}{dx} givens you the gradients of the tangents to the curve, so using implicit differentiation (let me know if you need me to explain this concept, i'll assume you know it already to save time):

    Spoiler:
    Show


    2x + 2y\frac{dy}{dx} = 0

    2y\frac{dy}{dx} = -2x

    \frac{dy}{dx} = \frac{-2x}{2y}

    \frac{dy}{dx} = \frac{-x}{y}

    subbing in the values x = 2 and y = 11 you get

    \frac{dy}{dx} = \frac{-2}{11}

    now using the formula y - y_1 = m(x - x_1) to find an equation of a line we get

    y - 11 = \frac{-2}{11}(x - 2)


    y = \frac{-2x}{11} + \frac{4}{11} + 11

    \therefore y = \frac{-2x}{11} + \frac{125}{11} or [latex]11y + 2x - 125 = 0

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    (Original post by CCDB)
    2) SHOW THAT THE STRAIGHT LINE

    x - 3y - 10 = 0

    IS A TANGENT TO THE CIRCLE

    x^2 + y^2 = 10

    BY: (I) FINDING THE DISTANCE OF THE LINE FROM THE ORIGIN
    (II) SHOWING THAT THE LINE MEETS THE CIRCLE AT ONE POINT ONLY

    FURTHER, DETERMINE THE EQUATION TO THE NORMAL TO THE CIRCLE THROUGH THE POINT OF CONTACT OF THE TANGENT
    part i

    Spoiler:
    Show

    the shortest distance from the line to the origin will be perpendicular to the line, therefore this line will have a gradient of -3 (product of gradients of perpendicular lines is -1) also this line must pass the origin. \therefore the equation of the line is y = -3x and if we equate this to the original line we get x + 9x - 10 = 0 \therefore x = 1 subbing into the original line we can find y. we get y = -3 using pythagorus we can find the shortest distance from the origin which is \sqrt{1^2 + (-3)^2} = \sqrt{10} this is the same as the radius of the circle



    part ii

    Spoiler:
    Show


    equate the lines to find any points of intersect.

    x^2 + y^2 = 10 (eq 1) and x = 10 + 3y (eq 2) (rearranged)

    subbing x from eq 2 into eq 1 we get

    (10 + 3y)^2 + y^2 = 10

    100 + 60y + 9y^2 + y^2 = 10

    10 + 6y + y^2 = 1

    y^2 + 6y + 9 = 0

    (y + 3)^2 = 0

    \therefore y = -3 and x = 1 (to get this sub into eq 2 or 1)

    so theres only one point of intersect \therefore it is a tangent

 
 
 
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