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

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Answer 1 · 2015-07-08T00:47:39

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

$(x-4)^2 + (y-5)^2 < 9$

This is the standard form for the equation of a circle with centre at ( $4,5$ ) and radius $sqrt9 = 3$ .

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]}

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