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

Question

mathmate · Answer

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 · Answer

since 3,-4 is the upper part of a circle; sqrt(yada)

amistre64 · Answer

3 over 4 down, doh!!  under side

amistre64 · Answer

-sqrt(yada)

anonymous · Answer

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 · Answer

now point slope formula gives it

anonymous · Answer

hmmm

anonymous · Answer

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 · Answer

but I'm in the parametric equations of line section? we haven't been using any derivatives

mathmate · Answer

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.