you can simplify them to t = root3s using the two equations you have given...

can we see the complete question?

Cheers

First of all, if 2 lines are perpendicular then the product of their slopes is -1.

Since both lines cross thru centre (0,0);

Slope 1 = (1 - 0) / (-root3 - 0) = - 1 / root3

Slope 2 = t - 0 / s - 0 = t/s

product of slopes = t/s * -1/root3 = -1 ==> t = root3 s

Now lets use the triangles;

r^2 = 1^2 + (-root3)^2 = 1 + 3 = 4

radius r = 2

apply this the other triangle;

r^2 = s^2 + t^2

4 = s^2 + (root3 s)^2 = s^2 + 3 s^2 = 4 s^2

2 = 2s ==> s = 1

Cheers

Thank you for the response.
The slope never struck my mind.

There is a little-known shortcut for this kind of a problem. Whenever you have a set-up such as this, the opposite side will simply be flipped. In other words, since the coordinates of one side are x = -sqroot3, and y = 1, then the other side will be x = 1 and y = sqroot3.

It's similar to the "testing points" approach on a line. If you know that points less than a given coordinate are negative, then the testing points will be in the form negative, positive, negative.

AleksandrM - Could you please explain the "touch point" formula please with an example?

Looking at the co-ordinates of P, we can deduce that OP=2 (Draw a line perpendicular to the x-axis from P to point B and use the pythogoras theorem).
If OP=2, PB=1 and OB=|root 3| we can deduce triangle OPB to be a 30-60-90 triangle with angle POB = 30.
Draw a perpendicular line from Q to the x-axis to Point A. So Angle QOA = 180 - (90+30)= 60.
Hence Triangle QOA is also 30-60-90 with hypotenuse=radius=2
Hence s= side opp to angle 30 degrees = 1/2 hypotenuse = 1
s=1, t= root3

smohsin wrote:AleksandrM - Could you please explain the "touch point" formula please with an example?

Sorry. I shouldn't have included that. I was just making a comparison to make a point of how shortcuts can sometimes be useful.

Testing points is from geometry. You apply it to inequalities. Don't worry about it, has nothing to do with this problem.