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Math Help - Finding the line tangent to a point on a circle...

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    Finding the line tangent to a point on a circle...

    Hi guys!
    I was out sick for the last week or so and am catching up on all the calculus homework that I missed. There's one in particular that I'm having trouble with. The problem is to find the slope-intercept form of the line tangent to the point (6,8) on the circle x^2 + y^2 = 100. I know that the slope of a secant line is [f(x+h) - f(x)]/h and since x is 6 I changed the expression to this: [f(h+6) - 8]/h. I know that f(x) = sqrt(100 - x^2) so f(6+h) comes out to sqrt(64 - 12x - h^2) so [f(h+6) - 8]/h turns into [sqrt(64 - 12x - h^2) -8]/h and I'm not sure where to go from there since I can't seem to get h out of the denominator so I can set h to 0 without getting an undefined answer. Any help would be really appreciated. Thanks in advance!
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    Re: Finding the line tangent to a point on a circle...

    Quote Originally Posted by imanironmaiden View Post
    The problem is to find the slope-intercept form of the line tangent to the point (6,8) on the circle x^2 + y^2 = 100.
    The derivative is 2x+2yy^{\prime}=0 so y^{\prime}=\frac{-x}{y}.
    So use (6,8) to find the slope y^{\prime}.
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    Re: Finding the line tangent to a point on a circle...

    Thanks Plato! My class hasn't touched on laws of derivatives, etc. so at this point your response looks like magic to me. Using the method I tried to use, how would I arrive at that answer and if it's not possible, why not? Thanks again for your answer!
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    Re: Finding the line tangent to a point on a circle...

    Quote Originally Posted by imanironmaiden View Post
    Hi guys!
    I was out sick for the last week or so and am catching up on all the calculus homework that I missed. There's one in particular that I'm having trouble with. The problem is to find the slope-intercept form of the line tangent to the point (6,8) on the circle x^2 + y^2 = 100. I know that the slope of a secant line is [f(x+h) - f(x)]/h and since x is 6 I changed the expression to this: [f(h+6) - 8]/h. I know that f(x) = sqrt(100 - x^2) so f(6+h) comes out to sqrt(64 - 12x - h^2) so [f(h+6) - 8]/h turns into [sqrt(64 - 12x - h^2) -8]/h and I'm not sure where to go from there since I can't seem to get h out of the denominator so I can set h to 0 without getting an undefined answer.
    I my view this a busy work exercise. But here it is.

    \frac{{\sqrt {100 - \left( {{x^2} + 2hx + {h^2}} \right)}  - \sqrt {100 - {x^2}} }}{h} = \frac{{ - 2hx - {h^2}}}{{h\left( {\sqrt {100 - \left( {{x^2} + 2hx + {h^2}} \right)}  + \sqrt {100 - {x^2}} } \right)}}

    The h's divide off.
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    Re: Finding the line tangent to a point on a circle...

    The tangent to a circle, at a given point, is always perpendicular to the radius to that point. If the given point is (x_0, y_0) (and, of course, x_0^2+ y_0^2= 100) then the line from the center to that point, from (0, 0) to (x_0, y_0) has slope \frac{y_0}{x_0}. Therefore, the slope of the tangent line is -\frac{x_0}{y_0}.
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    Re: Finding the line tangent to a point on a circle...

    Another pre-calculus technique is to let the tangent line be:

    y=mx+b

    Now, substitute for y into the equation of the circle:

    x^2+(mx+b)^2=100

    x^2+m^2x^2+2bmx+b^2=100

    \left(m^2+1 \right)x^2+(2bm)x+\left(b^2-100 \right)=0

    In order for the line to be tangent to the circle, we require 1 real root, hence the discriminant must be zero:

    (2bm)^2-4\left(m^2+1 \right)\left(b^2-100 \right)=0

    4b^2m^2-4\left(b^2m^2-100m^2+b^2-100 \right)=0

    400m^2-4b^2+400=0

    100m^2-b^2+100=0

    Now, given the point (6,8) is on the line, we know:

    b=8-6m

    100m^2-(8-6m)^2+100=0

    100m^2-64+96m-36m^2+100=0

    64m^2+96m+36=0

    16m^2+24m+9=0

    (4m+3)^2=0

    m=-\frac{3}{4}

    b=\frac{25}{2}

    Hence, the slope-intercept form of the tangent line is:

    y=-\frac{3}{4}x+\frac{25}{2}
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