# How do I solve this quadratic equation in terms of t?

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:

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

• Put the square root terms on one side, everything else on the other, and square. – Chappers Jul 25 '17 at 16:29
• @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. – name Jul 25 '17 at 16:31
• It's probably the only way, without reparametrising using $t=\sin{\theta}$ or something. – Chappers Jul 25 '17 at 16:39
• what do we know about the variables $$a,b,c,d$$? – Dr. Sonnhard Graubner Jul 25 '17 at 16:43
• @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. – 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?
• 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. – name 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 $\overline {PQ}$.