anonymous
  • anonymous
Find the line that is tangent to the circle x^2+y^2=25 at the point 3,-4.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmate
  • mathmate
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)
amistre64
  • amistre64
since 3,-4 is the upper part of a circle; sqrt(yada)
amistre64
  • amistre64
3 over 4 down, doh!! under side

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amistre64
  • amistre64
-sqrt(yada)
anonymous
  • anonymous
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 \[-\frac{3}{-4}=\frac{3}{4}\]
anonymous
  • anonymous
now point slope formula gives it
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
that was implicit diff that i used. you can solve for y if you like and then take the derivative there. you will get \[y'=\pm\frac{x}{\sqrt{1-x^2}}\]
anonymous
  • anonymous
but I'm in the parametric equations of line section? we haven't been using any derivatives
mathmate
  • mathmate
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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