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anonymous

  • 5 years ago

Find the line that is tangent to the circle x^2+y^2=25 at the point 3,-4.

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  1. mathmate
    • 5 years ago
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    The centre is at (0,0), the radial line passing through the centre (origin) has a slope of (-4/3), so the slope of the tangent is 3/4. A line through (3,-4) with slope 3/4 is therefore y=(3/4)(x-3)+(-4)

  2. amistre64
    • 5 years ago
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    since 3,-4 is the upper part of a circle; sqrt(yada)

  3. amistre64
    • 5 years ago
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    3 over 4 down, doh!! under side

  4. amistre64
    • 5 years ago
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    -sqrt(yada)

  5. anonymous
    • 5 years ago
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    and how do we know what the tangent slope is? start with \[x^2+y^2=25\] then \[2x+2yy'=0\] \[y'=-\frac{x}{y}\] and so at \[(3,-4)\] the slope is \[-\frac{3}{-4}=\frac{3}{4}\]

  6. anonymous
    • 5 years ago
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    now point slope formula gives it

  7. anonymous
    • 5 years ago
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    hmmm

  8. anonymous
    • 5 years ago
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    that was implicit diff that i used. you can solve for y if you like and then take the derivative there. you will get \[y'=\pm\frac{x}{\sqrt{1-x^2}}\]

  9. anonymous
    • 5 years ago
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    but I'm in the parametric equations of line section? we haven't been using any derivatives

  10. mathmate
    • 5 years ago
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    You don't need any derivative to find the slope of the radial line and the tangent line. You only need the relationship m1*m2 = -1. ] Follow the reasoning of my very first response.

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