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?


Let $O(1,0)$ be the center of the given circle $(x-1)^2 + y^2 =5$. Then,
$$PO^2 = (4-1)^2+(3-0)^2 = 18, \>\>\>\>\>OA^2 = 5$$
Use the fact that PAO is a right triangle to calculate,
$$PA = \sqrt{PO^2-OA^2} = \sqrt{18-5}=\sqrt{13}$$
