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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.
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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mathmate
 5 years ago
Best ResponseYou've already chosen the best response.0The 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)(x3)+(4)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since 3,4 is the upper part of a circle; sqrt(yada)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03 over 4 down, doh!! under side

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0now point slope formula gives it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that was implicit diff that i used. you can solve for y if you like and then take the derivative there. you will get \[y'=\pm\frac{x}{\sqrt{1x^2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but I'm in the parametric equations of line section? we haven't been using any derivatives

mathmate
 5 years ago
Best ResponseYou've already chosen the best response.0You 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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