Find the equation of tangent lines that go through point P((3,-5) to the circumference
x^2+y^2-x+3y-2=0

Question

Find the equation of tangent lines that go through point P((3,-5) to the circumference 
x^2+y^2-x+3y-2=0

anonymous · Answer

Find the derivative using implicit differentiation
$$

2 x + 2 y y' -1 + 3 y'=0\
2 y y' + 3 y'= 1-2 x  \
y'( 2y +3) = 1-2 x \
y'= \frac { 1-2 x}{( 2y +3)}
$$

anonymous · Answer

Find y' at the point P((3,-5) 
$$
y'= \frac {1 -6}   {-10+3}=\frac{-5}    {-7}=\frac 5 7
$$

anonymous · Answer

Now use

$$
y-y_0 = \frac 5 7 (x-x0)

$$

anonymous · Answer

At (3,-5)

$$y-y_0 = \frac 5 7 (x-x0)\
y+5 =\frac 5 7 (x-3)\
y  =-5+\frac 5 7 (x-3)
$$

anonymous · Answer

Did you get it?

owlcoffee · Answer

yeah, I got curious about that implicit differentiation.
What does it mean geometrically?

anonymous · Answer

To be able to find the equation of the tangent, you have to find the derivative.