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Math Help - Coordinate Geometry of Circles

  1. #1
    Member maybeline9216's Avatar
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    Coordinate Geometry of Circles

    The equation of a circle, C is x^2+y^2-8x+10y+28=0

    (i) Find the coords. of the centre of C and find the radius of C.

    (ii)Show that the line L:2x+3y=6 is a tangent to the circle and find the coords. of the point of contact.

    Please do part(ii). Anyway i have found the coords. of the pt. of contact which is (6,-2). But how to show that the line is a tangent to the circle??
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    Quote Originally Posted by maybeline9216 View Post
    The equation of a circle, C is x^2+y^2-8x+10y+28=0

    (i) Find the coords. of the centre of C and find the radius of C.

    (ii)Show that the line L:2x+3y=6 is a tangent to the circle and find the coords. of the point of contact.

    Please do part(ii). Anyway i have found the coords. of the pt. of contact which is (6,-2). But how to show that the line is a tangent to the circle??
    The line from point of contact to centre is radius. If this line (radius) is perpendicular to the given line 2x + 3y =6, then this given line is tangent.

    Now
    Centre = (4, -5)
    Point of contact=(6, -2)

    slope of radius = \frac{-2-(-5)}{6-4}=\frac{3}{2}

    Slope of given line 2x + 3y = 6 is = \frac{-2}{3}

    Product of their slopes = \frac{3}{2} \times \frac{-2}{3} = -1

    So, the given line is perpendicular to the radius at the point of contact.
    so, it is a tangent at rhe point of contact.
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