# Circle A has a center at #(12 ,9 )# and an area of #16 pi#. Circle B has a center at #(3 ,1 )# and an area of #67 pi#. Do the circles overlap?

Yes.

First we need to find the radii of the circles. We can do this using the formula for area:

Circle A

Circle B

We now find the distance between the centres. We can use the distance formula for this:

Coordinates

Let

and

If:

Sum of radii:

This shows the circles intersect at two points.

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To determine if the circles overlap, we need to calculate the distance between their centers and compare it to the sum of their radii. If the distance between the centers is less than the sum of their radii, then the circles overlap.

Let's denote the centers of Circle A and Circle B as ( (x_1, y_1) ) and ( (x_2, y_2) ) respectively, and their respective radii as ( r_1 ) and ( r_2 ).

For Circle A: Center: ( (12, 9) ) Radius: ( r_1 = \sqrt{\frac{16\pi}{\pi}} = 4 )

For Circle B: Center: ( (3, 1) ) Radius: ( r_2 = \sqrt{\frac{67\pi}{\pi}} = \sqrt{67} )

Now, let's calculate the distance between the centers using the distance formula:

[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

[ \text{Distance} = \sqrt{(3 - 12)^2 + (1 - 9)^2} ]

[ \text{Distance} = \sqrt{(-9)^2 + (-8)^2} ]

[ \text{Distance} = \sqrt{81 + 64} ]

[ \text{Distance} = \sqrt{145} ]

Now, we compare the distance between the centers to the sum of their radii:

[ \text{Distance} < r_1 + r_2 ]

[ \sqrt{145} < 4 + \sqrt{67} ]

Since ( \sqrt{145} ) is less than ( 4 + \sqrt{67} ), the circles do overlap.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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