Circle Geometry Question 1) In triangle $ABC$, $AB = 10$, $AC = 8$, and $BC = 6$. Let $P$ be the point on the circumcircle of triangle $ABC$ so that $\angle PCA = 45^\circ$. Find $CP$.
Diagram(1)
2) Let $B$, $C$, and $D$ be points on a circle. Let $\overline{BC}$ and the tangent to the circle at $D$ intersect at $A$. If $AB = 4$, $AD = 8$, and $\overline{AC} \perp \overline{AD}$, then find $CD$. Diagram(2)
I think both of them have to do with power of a point 
 A: Analytical solution for question (1):
Take coordinate axes centered in $C$ with natural coordinate axes directed by $\vec{CA}$ and $\vec{CB}$ giving the following coordinates: $A(8,0)$ and $B(0,6)$. 
Let $(C)$ be the circle. Its center $\Omega$ clearly has coordinates $(4,3)$. 
Thus the equation of $(C)$ has the form 
$$(x-4)^2+(y-3)^2=R^2 \ \ \ (*)$$
Knowing that $(x,y)=(0,0)$ is a solution of (*), $R^2=25$. Therefore, the equation of (C) is:
$$x^2+y^2-8x-6y=0 \ \ \ (1) $$
The abscissas of the intersection points of $(C)$ with straight line 
$$y=x \ \ \ (2)$$ 
($\pi/4$ angle means slope 1).
are obtained by plugging (2) into (1), giving $x(x-7)=0$. Excluding $x=0$ that corresponds to $C$, it remains $x=7$ for the abscissa of $P$. Thus $P(7,7)$. 
Therefore $CP=7\sqrt{2}$.
A: 1) Since the angle $\angle PCA = 45^\circ$ and the $\triangle ACB$ is clearly Pythagorean (double of the minimal of sides $3,4,5$) the diameter of the circle is equal to 
$\overline{AB}$ and because the angle $\angle APB$ subtends an arc of $90^\circ$ the quadrilatere $ACBP$ is a rectangle. Hence $$\overline{CP}=\overline{AB}=\color{red}{10}$$
2) By the power of the point $A$ respect to the given circle we have $$4(4+x)=8^2\iff x=12$$ It follows 
$$\overline{CD}=\sqrt{8^2+16^2}=\color{red}{8\sqrt 5}$$
