Where Are the Tangent Points on the Circle?

[Chris L T521]

Here's this week's problem!

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Problem: Find two points on the circle $x^2+(y-1)^2=1$ at which a level curve of $f(x,y)=x^2-y^2$ is tangent to the circle.

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Suggestion: [sp]The first equation's graph is a level curve of $g(x,y)=x^2+(y-1)^2-1$; at a point of tangency the gradients of $f$ and $g$ will be parallel ($\nabla f = \lambda \nabla g$ for some number $\lambda$). Solve the resulting system of three equations in the three unknowns $x$, $y$, and $\lambda$.[/sp]

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

No one answered this week's problem. I will post a solution for this soon (hopefully tomorrow); delays are due to having a mentally draining job and spending a lot of my times these days doing GRE prep for another round of applications to PhD programs this fall.