# Prove that there is a unique circle...

• Jan 16th 2013, 03:13 PM
TimsBobby2
Prove that there is a unique circle...
Let A be a point on a line l, and let B be a point that does not lie on l. Prove that there is a unique circle that contains A and B and is tangent to l.

This is obviously true, but I am getting nowhere. Can I have any help please?
• Jan 16th 2013, 03:43 PM
Plato
Re: Prove that there is a unique circle...
Quote:

Originally Posted by TimsBobby2
Let A be a point on a line l, and let B be a point that does not lie on l. Prove that there is a unique circle that contains A and B and is tangent to l.

There a unique ray $\displaystyle \overrightarrow {AP}$ that is perpendicular to $\displaystyle \ell$ that is in the $\displaystyle B\text{-side}\cup\ell.$

Let $\displaystyle \alpha$ be the perpendicular bisector of $\displaystyle \overline {AB}$.

Can show that $\displaystyle \alpha\cap\overrightarrow {AP}\ne\emptyset~?$

Is that point the center of the circle?
• Jan 16th 2013, 03:59 PM
TimsBobby2
Re: Prove that there is a unique circle...
What do you mean the B side unioned with l? That's more confusing to me really, could you be a little more clear?
• Jan 16th 2013, 04:11 PM
Plato
Re: Prove that there is a unique circle...
Quote:

Originally Posted by TimsBobby2
What do you mean the B side unioned with l? That's more confusing to me really, could you be a little more clear?

I assumed you were in an axiomatic geometry course.
But I guess you are not.

There is a unique plane, $\displaystyle \Pi$, determined by $\displaystyle \ell~\&~B$

Look at the unique perpendicular to $\displaystyle \ell$ at $\displaystyle A$ that is in $\displaystyle \Pi$.

That perpendicular will intersect the perpendicular bisector of $\displaystyle \overline {AB}$.

That is the center of your circle.