Question

If a line passing through the origin touches the circle ${(x-4)}^2+{(y-5)}^2=25$, then find its slope.

Answer 1

we get the slope by differentiating the equation of circle

$\frac{dy}{dx}=slope$

$\frac{d}{dx}\left(\right(x-4{)}^2)+(y-5{)}^2-25)=0$

$=2(x-4)+2(y-5)\frac{dy}{dx}=0$

given line passes through origin that is put $x=0,y=0$

$-8-10\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-4}{5}=slope$