I'm having trouble with this equation. I thought it would be easy to solve using the quadratic formula, but I have no idea how to start. This is not homework, but I would like to know how to solve this in terms of $t$. Both sides are pretty much the same since it has to do with two points $(a,b)$ and $(c,d)$ which are equidistant to the point $(t,\sqrt{1-t^2})$, but this information is not relevant to the problem. How do I isolate $t$ in this situation:

$a^2-2b\cdot \sqrt{1-t^2}+b^2+2ct=c^2-2d\cdot \sqrt{1-t^2}+d^2+2at$

  • $\begingroup$ Put the square root terms on one side, everything else on the other, and square. $\endgroup$ – Chappers Jul 25 '17 at 16:29
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    $\begingroup$ @Chappers That would involve a lot of distributing wouldn't it? If that is the best way to solve it then I guess I'm up for the work. $\endgroup$ – name Jul 25 '17 at 16:31
  • $\begingroup$ It's probably the only way, without reparametrising using $t=\sin{\theta}$ or something. $\endgroup$ – Chappers Jul 25 '17 at 16:39
  • $\begingroup$ what do we know about the variables $$a,b,c,d$$? $\endgroup$ – Dr. Sonnhard Graubner Jul 25 '17 at 16:43
  • $\begingroup$ @Dr.SonnhardGraubner Only that the points $(a,b)$ and $(c,d)$ lie on the same circle that the point $(t, \sqrt{1-t^2})$ lies on. Essentially we know nothing about them, since the equation is for a generalization. $\endgroup$ – name Jul 25 '17 at 16:47

HINT: your equation can be written in the form $$a^2+b^2-c^2-d^2+(2c-2a)t=\sqrt{1-t^2}(2b-2c)$$ we denote by $$A=a^2+b^2-c^2-d^2$$ $$B=(2c-2a)$$ $$C=2b-2d$$ then we have $$A+Bt=\sqrt{1-t^2}C$$ after squaring we get $$t^2(B^2+C^2)-2ABt+A^2-C^2=0$$ can you solve this?

  • $\begingroup$ Wow! Your use of extra variables really help. That is way easier, thank you. I think the quadratic formula can be used directly now. $\endgroup$ – name Jul 25 '17 at 16:53
  • $\begingroup$ or should i post the solution for you? $\endgroup$ – Dr. Sonnhard Graubner Jul 25 '17 at 16:56
  • $\begingroup$ I solved it thank you. For the test points $(1, 0)$ and $(0, 1)$ as $(a,b)$ and $(c,d)$ respectively I found that $t=0.7071...$ as expected. $\endgroup$ – name Jul 25 '17 at 17:02
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    $\begingroup$ so is all ok? that is fine $\endgroup$ – Dr. Sonnhard Graubner Jul 25 '17 at 17:02

The solution of your problem are the points of the unit circle $u$ that are equidistant to the points $P (a,b)$ and $Q (c,d)$. These can be found by intersecting $u$ with the line segment bisector of the segment $\overline {PQ}$.

  • $\begingroup$ I didn't think about it that way. That's even easier. $\endgroup$ – name Jul 25 '17 at 21:32
  • $\begingroup$ So technically, you could continuously bisect more and more angles on the circle until you get closer to a desired angle, and then be able to find the sine, cosine, and tangent of that angle (based on the point on the circle). Is this an actual feasible way to calculate trig values? And does this method actually have a name? $\endgroup$ – name Jul 25 '17 at 21:42

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