1. ## Circles

In circle O and circle P, AP is equal to 10 and AC is equal to 8. Line segment CB is also perpendicular to line segment AP. Find the measure of DP.

Here's what I have so far : Since all radii of a circle are congruent, triangle CBA is an isosceles triangle. And since base angles of an isosceles triangle are congruent, angle C and angle A are both equal to 45. After there, I guess I could use trig to figure out the measures of CB and go from there, but I think I'm going in the wrong direction.
Can anybody help me?

2. It's wrong.
As the picture looks, $CB$ is not a radius of circle O.
In triangle ACP we have $CP^2=AP^2-AC^2=36$.
Then applying cathetus' theorem we have
$CP^2=AP\cdot PB\Rightarrow PB=\frac{CP^2}{AP}=\frac{18}{5}$
So, $DP=BP=\frac{18}{5}$

3. Originally Posted by blindspot
In circle O and circle P, AP is equal to 10 and AC is equal to 8. Line segment CB is also perpendicular to line segment AP. Find the measure of DP.

Here's what I have so far : Since all radii of a circle are congruent, triangle CBA is an isosceles triangle.
BC doesn't start or end at O so it is not a radius.

If you are allowed to use trigonometry, here's a start.

Connect CP. Then triangle ACP is a right triangle with angle ACP the right angle. So angle BAC (call this $\theta$) can be found by:
$cos( \theta) = \frac{8}{10}$

Triangle ABC is also a right triangle, so we know that
$BC = 8 \cdot sin(\theta) = 8 \sqrt{1 - cos^2(\theta)} = 8 \cdot \sqrt{1 - \left ( \frac{8}{10} \right )^2}$ $= 8 \cdot \sqrt{\frac{36}{100}} = 8 \cdot \frac{6}{10} = 4.8$

Can you take it from here?

(Another hint: angle PCB is also $\theta$.)

-Dan

4. red_dog: Thanks for the help, but I haven't learned cathetus' theorem, so I can't use that.
topsquark: If PCB is also equal to a theta, then does that mean I have to redo the work that you did for BC to find the measure of BP? Then would I draw axillary line DO and use the Pythagorean Theorem? But to be able to use that, I would have to prove that triangle BPD is a right triangle.
Sorry, but could you help me some more?

(I'm not sure if you can though since, you appeared to have signed off)

5. Originally Posted by blindspot
red_dog: Thanks for the help, but I haven't learned cathetus' theorem, so I can't use that.
topsquark: If PCB is also equal to a theta, then does that mean I have to redo the work that you did for BC to find the measure of BP? Then would I draw axillary line DO and use the Pythagorean Theorem? But to be able to use that, I would have to prove that triangle BPD is a right triangle.
Sorry, but could you help me some more?

(I'm not sure if you can though since, you appeared to have signed off)
Triangle BCP is also a right triangle with the angle BCP equal to $\theta$. You know how to find the value of $sin(\theta)$. You are trying to find BP. Does that give you any ideas?

-Dan

6. I think what you're trying to tell me is that:

$BP = 8 \cdot sin(\theta) = 8 \sqrt{1 - cos^2(\theta)} = 8 \cdot \sqrt{1 - \left ( \frac{8}{10} \right )^2}$ $= 8 \cdot \sqrt{\frac{36}{100}} = 8 \cdot \frac{6}{10} = 4.8$

Does that mean that DP is equal to 4.8? (Since I think that they're both radii and radii of a circle are congruent)

7. Originally Posted by blindspot
I think what you're trying to tell me is that:

$BP = 8 \cdot sin(\theta) = 8 \sqrt{1 - cos^2(\theta)} = 8 \cdot \sqrt{1 - \left ( \frac{8}{10} \right )^2}$ $= 8 \cdot \sqrt{\frac{36}{100}} = 8 \cdot \frac{6}{10} = 4.8$

Does that mean that DP is equal to 4.8? (Since I think that they're both radii and radii of a circle are congruent)

Hmmm... I wasn't incorrect in my calculations, but I think I may have mislead you. I really didn't need you to calculate BC.

What's the length of the new line CP? (Look at triangle ACP.) It's 6. We know the sine of angle CBP and you know the hypotenuse (CP) of the right triangle BCP, so you can find the leg BP. And since BP = PD...

-Dan

8. Hello, blindspot!

In circle $O$ and circle $P\!,\;AP = 10$ and $AC = 8.$
Also: . $CB \perp AP.$
Find the measure of $DP.$

Let $BP = DP = r$, then $AB = 10-r$
Let $h = CB$.

Since $\Delta ACB$ is inscribed in a semicircle,
. . it is a right triangle with $\angle ACB = 90^o$.
We have: . $CP^2 + AC^2 \:=\:AP^2\quad\Rightarrow\quad CP^2 + 8^2 \:=\:10^2\quad\Rightarrow\quad CP \,=\,6$

Draw $CP$.
In right triangle $CBP\!:\;\;r^2 + h^2\:=\:6^2$ .[1]

Since $h$ is the altitude to the hypotenuse of a right triangle,
. . it is the mean proportional between the two segments of the hypotenuse.

Hence, we have: . $h^2\:=\:r(10-r)\quad\Rightarrow\quad h^2\:=\:10r - r^2$

And we have: . $10r \:=\:r^2 + h^2$ .[2]

Substitute [1] into [2]: . $10r \:=\:36\quad\Rightarrow\quad r \,=\,DP\,=\,3.6$