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

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

Mathematics
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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)
since 3,-4 is the upper part of a circle; sqrt(yada)
3 over 4 down, doh!! under side

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Other answers:

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