hyperbolic geometry (help!) angles.

**gomes** · May 31st 2010, 01:46 PM

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

Cheers.

**Bruno J.** · May 31st 2010, 11:27 PM

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.

**HallsofIvy** · June 1st 2010, 01:51 AM

$|z+1|= 2\sqrt{2}$ 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.

**gomes** · June 1st 2010, 10:21 PM

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.

**gomes** · June 2nd 2010, 12:50 AM

whoops, made a slight mistake..

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

**gomes** · June 2nd 2010, 12:53 AM

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

**HallsofIvy** · June 2nd 2010, 03:23 AM

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.

**gomes** · June 2nd 2010, 03:25 AM

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

**gomes** · June 4th 2010, 06:46 AM

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?

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