How do you graph (x-4)^2 + (y-5)^2 < 9?

Jul 8, 2015

Pretend that the inequality is an equality, solve the equation and plot the graph. Shade all areas that satisfy the inequality.

Explanation:

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} < 9$

This is the standard form for the equation of a circle with centre at ($4 , 5$) and radius $\sqrt{9} = 3$.

This means that, from the centre, you go 3 units to the right, 3 to the left, 3 up, and 3 down.

Thus, the four extreme points are at ($4 , 8$), ($4 , 2$) ($1 , 5$), and ($7 , 5$).

Since the inequality is "< 9", you use a dotted line for the graph, and you shade all areas for which $y < 9$.

graph{(x-4)^2 + (y-5)^2 < 9 [-6.41, 13.59, -1.04, 8.96]}