# 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

• In the first one, $AB$ is the diameter and $\angle PBA =\angle PAB = 45 ^\circ$. Then you can use Ptolemy's theorem. Jul 9 '16 at 20:38
• In the second one, power of point, yes. $4AC=8^2$, and $AC^2+8^2=CD^2$ by Pythagoras. Jul 9 '16 at 20:43

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 $CP=7\sqrt{2}$.

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 $$\overline{CD}=\sqrt{8^2+16^2}=\color{red}{8\sqrt 5}$$