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Math Help - [SOLVED] equation of a tangent to circle

  1. #1
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    [SOLVED] equation of a tangent to circle

    last one now honest

    a circle has the equation x^2+y^2-8x-4y=9


    centre is C (4, 2)

    AB is diameter A (2,7) and B (6, -3)

    find equation of the tangent to the circle at A. give your answer in the form y=mx+c

    thanks!
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  2. #2
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    Quote Originally Posted by coyoteflare View Post
    last one now honest

    a circle has the equation x^2+y^2-8x-4y=9


    centre is C (4, 2)

    AB is diameter A (2,7) and B (6, -3)

    find equation of the tangent to the circle at A. give your answer in the form y=mx+c

    thanks!
    Hello again coyote,

    First, find the slope of line BA.

    m=\frac{7-^-3}{2-6}=-\frac{5}{2}

    The tangent line (t) through A(2, 7) will be perpendicular to BA.

    The slope of the tangent line (t) will be the negative reciprocal of the slope of BA or \frac{2}{5}

    Now, all you need to do is determine the equation of the line that passes through A(2, 7) with a slope of \frac{2}{5}.

    Use y=mx+c, x = 2, y = 7, and m = \frac{2}{5}

    Solve for c

    7=\frac{2}{5}(2)+c

    Can you complete it?
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  3. #3
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    Quote Originally Posted by coyoteflare View Post
    last one now honest

    a circle has the equation x^2+y^2-8x-4y=9


    centre is C (4, 2)

    AB is diameter A (2,7) and B (6, -3)

    find equation of the tangent to the circle at A. give your answer in the form y=mx+c

    thanks!
    The line from (4, 2) to (2, 7), a radius of the circle, has slope (7- 2)/(2- 4)= -5/2. The tangent at that point must be perpendicular to the radius and so must have slope 2/5. The line through (2, 7) with slope 2/5 is y= (2/5)(x- 2)+ 7.

    Alternatively, differentiate the equation with respect to x: 2x+ 2y dy/dx- 8- 4 dy/dx= 0 so (2y- 4)dy/dx= 8- 2x. At x= 2, y= 7, that is (14- 4)dy/dx= 8-4 or dy/dx= 4/10= 2/5 just as before. Again, the line through (2, 7) with slope 2/5 is y= (2/5)(x- 2)+ 7.
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