Equation of the line given the tangent lines

Question

Tangent lines to

$x^2+y^2-18x-2y+48=0$

at points where $x = 4$

How do you solve this kind of problem? P.S. the answer to this problem is

$y = 5/3 x - 8/3$

and

$y = - 5/3 x + 14/3$

Answer 1

Take differentiation w.r.t. $x$ both sides, $$ 2x+2y\frac{dy}{dx}-18-2\frac{dy}{dx}+0=0 $$ $$ \frac{dy}{dx}=\frac{9-x}{y-1} $$ At $x=4$ equation $(x^2+y^2-18x-2y+48=0)$ becomes , $$ 16+y^2-72-2y+48=0 $$ $$ y^2-2y-8=0 $$ This gives $y=-2$ or $y=4$ so, $$ \frac{dy}{dx}=\frac{9-x}{y-1}=>(\frac{dy}{dx})_{x=4,y=-2}=\frac{-5}{3} $$ and $$ \frac{dy}{dx}=\frac{9-x}{y-1}=>(\frac{dy}{dx})_{x=4,y=4}=\frac{5}{3} $$ These two slopes at $(4,-2)$ and $(4,4)$ gives two lines as,

Line 1: $$ \frac{-5}{3}=\frac{y+2}{x-4} $$ On rearranging, $$ y = \frac{-5}{3} x + \frac{14}{3} $$ Line 2: $$ \frac{5}{3}=\frac{y-4}{x-4} $$ On rearranging, $$ y = \frac{5}{3} x - \frac{8}{3} $$