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Math Help - C2 Circle Co-ordinate Geometery

  1. #1
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    C2 Circle Co-ordinate Geometery

    A circle C has centre D and equation
    x + y + 2x – 8y + 8 = 0.
    (a) Find the coordinates of D and the radius of C. [3]
    (b) A line is drawn through the point P (4, 6) so that it touches the circle C at the point T.
    (i) Show that PT = √20.
    (ii) Find the equation of the circle centre P which passes through the point T. [5]

    I know that a] will give D (-1, 4) and a radius of 3
    But I have difficulty finding b and c. Can someone please help? I am having a lot of trouble with this and really need a bit of guidance.
    Any help is appreciated
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  2. #2
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    Quote Originally Posted by db5vry View Post
    A circle C has centre D and equation
    x + y + 2x 8y + 8 = 0.
    (a) Find the coordinates of D and the radius of C. [3]
    (b) A line is drawn through the point P (4, 6) so that it touches the circle C at the point T.
    (i) Show that PT = √20.
    (ii) Find the equation of the circle centre P which passes through the point T. [5]

    I know that a] will give D (-1, 4) and a radius of 3
    But I have difficulty finding b and c. Can someone please help? I am having a lot of trouble with this and really need a bit of guidance.
    Any help is appreciated
    Because a tangent is perpendicular to the radius in the tangent point the points D, P, T are the vertices of a right triangle with PD as hypotenuse:

    |\overline{PD}|=\sqrt{(4-(-1))^2 + (6-a)^2} = \sqrt{29}

    One leg of the right triangle is the radius with a length of 3. Use Pythagorean theorem:

    |\overline{PT}|=\sqrt{29-3^2}=\sqrt{20}

    The general equation of the circle with M(x_M, y_M) and radius r is:

    (x-x_M)^2 + (y-y_M)^2=r^2

    You know the coordinates of the centre and you know the length of the radius: r = |\overline{PT}|

    Therefore the equation of the circle is:

    (x-4)^2+(y-6)^2=20
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