# Thread: hyperbolic geometry (help!) angles.

1. ## hyperbolic geometry (help!) angles.

Im having a bit of trouble, im not sure how to solve this question? Whats the method?

Cheers.

2. A hyperbolic line is a semicircle perpendicular to the real line.

You just have to find the two intersection points of the two circles and keep the point with positive imaginary part.

3. $|z+1|= 2\sqrt{2}$ is a circle with center at (-1, 0) and radius $2\sqrt{2}$ and $|z-5|= 2\sqrt{5}$ is a circle with center at (5, 0) and radius $2\sqrt{5}$.

They are given by $(x+1)^2+ y^2= 8$ and $(x- 5)^2+ y^2= 20$. Subtracting one equation from the other gives a linear equation for x. After finding x, substitute back into either equation to get a quadratic equation for y. As Bruno J says, you want the point with positive y.

4. Originally Posted by Bruno J.
A hyperbolic line is a semicircle perpendicular to the real line.

You just have to find the two intersection points of the two circles and keep the point with positive imaginary part.
Thanks! what do you mean by "KEEP the point with positive imagine part"?

As in, i got x=-3, and y=+-4, or y=+-8i.

Originally Posted by HallsofIvy
$|z+1|= 2\sqrt{2}$ is a circle with center at (-1, 0) and radius $2\sqrt{2}$ and $|z-5|= 2\sqrt{5}$ is a circle with center at (5, 0) and radius $2\sqrt{5}$.

They are given by $(x+1)^2+ y^2= 8$ and $(x- 5)^2+ y^2= 20$. Subtracting one equation from the other gives a linear equation for x. After finding x, substitute back into either equation to get a quadratic equation for y. As Bruno J says, you want the point with positive y.
Thanks! please see above for what i've done. so i got x=-3, and y=+-4, or y=+-8i. How would i display my final answer?
2.

5. whoops, made a slight mistake..

So is the answer just z= 1 + 2i?

6. Sorry, here is the 2nd part, how would I solve this part? Whats the method?
(the lines are in the first post)

7. Originally Posted by gomes
Sorry, here is the 2nd part, how would I solve this part? Whats the method?
(the lines are in the first post)
The "Poincare' half plane model" preserves angles. Find the tangents to the two circles at the point of intersection and find the (Euclidean) angles between them.

8. wait, i think I got it now, let me try it.

9. Originally Posted by HallsofIvy
The "Poincare' half plane model" preserves angles. Find the tangents to the two circles at the point of intersection and find the (Euclidean) angles between them.
thanks, so i've differentiated both implicitly, how would i find the angle?