# Tuesday, Mar 21

In class today the lesson focused on the relationship between the Pythagorean Theorem and the equation for a circle. Students relate the distance between $\dpi{300}\inline (a, b)$ and $\dpi{300}\inline (x, y)$ to the pythagorean theorem, and realize that $\dpi{300}\inline d = \sqrt{(x - a)^2 + (y-b)^2}$ and then we note that the equation of a circle whose center is at $\dpi{300}\inline (a, b)$ and which has a radius of $\dpi{300}\inline r$ is

$\dpi{300}\inline r^2 = (x - a)^2 + (y-b)^2$

or

$\dpi{300}\inline r =\pm \sqrt{(x-a)^2 + (y-b)^2}$

Note, I just now realized that I did not include $\dpi{300}\inline \pm$ in the notes from class.

Handed back today: Chapter 8 Assessments on Criteria A and C. Students received two scores on the latest assessment.

Today's assignment: 9.5 on p. 489: 1-12

Upcoming: Assessment on Chapter 9 at end of week. See me if you plan to be absent on Thursday/Friday.