Let $$\mathbb{H} = \{(x,y) \in \mathbb{R^2} \ | \ y>0\}$$ $$_{a}L = \{(x,y) \in \mathbb{H} \ | \ x=a\}$$ $$_{c}L_r = \{(x,y) \in \mathbb{H} \ | \ (x-c)^2+y^2=r^2\}.$$

Prove that if $P$ and $Q$ are distinct points in $\mathbb{H}$ then they cannot lie simultaneously on both $_{a}L$ and $_{c}L_r$.

I can see intuitively why this is true with the restriction of $y>0$. However, this is not true with the full circle since both $_{c}L_r$ and $_{a}L$ can lie simultaneously without the restriction of $y>0$.

I proved it by contradiction.

Let $P = (x_1,y_1)$, $Q=(x_2,y_2)$ be distinct points where $y_1,y_2>0$. Suppose $P$ and $Q$ lie simultaneously on both $_{a}L$ and $_{c}L_r$ then $x_1=x_2=a$ but $c=\frac{y_2^2-y_1^2+x_2^2-x_1^2}{2(x_2-x_1)}$ hence $x_1\ne x_2$ thus a contradiction.

I don't think the way I proved it is correct since if $y$ doesn't have the restriction of $y>0$ then my proof will also make the full circle $(x-c)^2+y^2=r^2$ have the restriction of both $_{c}L_r$ and $_{a}L$ not being able to lie simultaneously.

  • $\begingroup$ Just because you wrote a formula equating $c$ with a fraction having a factor of $(x_2-x_1)$ in its denominator does not prove that $x_1 \ne x_2$. Perhaps your derivation of the formula for $c$ contains an illegal division by zero! $\endgroup$ – Lee Mosher Jan 28 '18 at 18:04
  • $\begingroup$ @LeeMosher ty for the clarification! $\endgroup$ – Robben Jan 28 '18 at 18:20

From $$(x_1 - c)^2 + y_1^2 = (x_2 - c)^2 + y_2^2$$ we obtain $$2(x_2 - x_1) c = y_2^2 - y_1^2 + x_2^2 - x_1^2$$ But $x_2 - x_1 = a - a = 0$, therefore we can't divide both sides by $x_2 - x_1$. Instead, since we also have that $x_2^2 - x_1^2 = a^2 - a^2 = 0$, we get $$y_2^2 - y_1^2 = 0$$ which implies that $y_2 = y_1$ or $y_2 = -y_1$.

If $y_2 = y_1$ then $P = Q$, which is a contradiction. If $y_2 = -y_1$ then $y_2 < 0$, which is again a contradiction.

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  • $\begingroup$ Awesome, thanks a bunch! $\endgroup$ – Robben Jan 28 '18 at 18:20

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