Distance from $P(4,3)$ to tengent point on curve $x^2 + y^2 -2x-4=0$ 
Tangent line to curve $$x^2 + y^2 -2x-4=0$$ at A passes $P(4,3)$.
Find distance A and P.

Tangent of the line is $\frac{2-2x}{\sqrt{2x+4-x^2}}$
A(k,l) means
$l = \sqrt{2k+4-k^2}$
And
$l = \frac{2-2k}{\sqrt{2k+4-k^2}}$
So, I get $k=\frac{3\pm\sqrt{21}}{2}$
$AP = \sqrt{(k-4)^2 + (l-3)^2}$
Its so complicated, am i wrong somewhere?

The required answer is $\sqrt{4^2+3^2-2•4-4}$ = $\sqrt{13}$
Use the formula $\sqrt{S1}$ Or proceed by using Pythagorean theorem as the other answer. 
