Question

Let $y=P$ be the tangent to the circle $x^2+y^2-2x-4y+k=0$, where $\frac{P}{3}=\underset{x\to 0}{lim}\frac{x-\tan x}{x^3}$, then the value of $|k|$ is

Answer 1

$\frac{P}{3}=\underset{x\to 0}{lim}\frac{x-\tan x}{x^3}$
Applying L'Hopital's Rule,
$=\underset{x\to 0}{lim}\frac{1-{\sec }^{2}x}{3x^2}=\underset{x\to 0}{lim}\frac{{\cos }^{2}x-1}{3x^2}=\underset{x\to 0}{lim}\frac{-{\sin }^{2}x}{3x^2}\Rightarrow \frac{P}{3}=\frac{-1}{3}\Rightarrow P=-1$

$\therefore \text{Equation of tangent:}y=-1$
$x^2+y^2-2x-4y+k=0\Rightarrow (x-1{)}^2+(y-2{)}^2=5-k$
Now, distance from center to tangent = radius
$\frac{|3|}{1}=\sqrt{5-k}\Rightarrow k=-4\Rightarrow |k|=4$