Which point is NOT on the circle with center (3, 2) and radius 5?

Mathematics

1 answer:

podryga [215]3 years ago

8 0

<h3>Answer: Choice A. (-2,0)</h3>

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Work Shown:

The center is (h,k) = (3,2) and the radius is r = 5

The standard circle equation (x-h)^2 + (y-k)^2 = r^2 turns into (x-3)^2+(y-2)^2 = 5^2 or (x-3)^2+(y-2)^2 = 25

The idea is to plug in each (x,y) point that is shown in the answer choices. Then simplify to see if you get a true equation or not. If you get a true equation, then that point is on the circle.

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Choice A

Plug in (x,y) = (-2,0)

(x-3)^2+(y-2)^2 = 25

(-2-3)^2+(0-2)^2 = 25

(-5)^2 + (-2)^2 = 25

25 + 4 = 25

29 = 25

We get a false equation, so we can stop here since we found the answer.

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I'll try out choice B to see if we get a true equation here or not.

(x-3)^2+(y-2)^2 = 25

(6-3)^2+(-2-2)^2 = 25 ... plug in (x,y) = (6,-2)

(3)^2 + (-4)^2 = 25

9 + 16 = 25

25 = 25

We get a true equation, so the point (6,-2) is on the circle. This means we can rule out choice B.

I'll let you try out the other two points. They should be on the circle, so you should get true equations after plugging in the coordinates. If you're still stuck, then let me know.

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