Circle A has a center at #(3 ,4 )# and an area of #18 pi#. Circle B has a center at #(8 ,1 )# and an area of #40 pi#. Do the circles overlap?

Answer 1

Circles Overlap

Distance between the centers of the circles using distance formula

#vec(O_AO_B) = sqrt((8-3)^2 + (1-4)^2) ~~ 5.83#

#A_A = pi (R_A)^2#

#R_A = sqrt((18pi) / pi ) ~~ 4.24#

Similarly #R_B = sqrt((40pi)/pi) ~~ 6.32#

Sum of the radii #R_A + R_B = 4.24 + 6.32 = 10.56#

Since #R_A + R_B > vec(O_AO_B)#, two circles overlap.

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Answer 2

To determine if the circles overlap, we need to compare the distances between the centers of the circles and the sum of their radii.

The distance between the centers of Circle A and Circle B can be calculated using the distance formula:

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Next, we need to calculate the radii of both circles using the formula:

Area=πr2Area = \pi r^2

Once we have the radii, we can compare the distance between the centers of the circles with the sum of their radii.

If the distance between the centers is greater than the sum of the radii, the circles do not overlap. If it is equal to or less than the sum of the radii, the circles overlap.

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Answer 3

To determine if the circles overlap, we can compare the distance between their centers to the sum of their radii. If the distance between the centers is less than the sum of their radii, the circles overlap. Otherwise, they do not overlap.

Let's denote the centers of the circles as (x1,y1) (x_1, y_1) for Circle A and (x2,y2) (x_2, y_2) for Circle B. The distance between the centers is given by the formula:

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

And the radii of the circles can be calculated as follows:

For Circle A, radius r1=A1π r_1 = \sqrt{\frac{A_1}{\pi}}

For Circle B, radius r2=A2π r_2 = \sqrt{\frac{A_2}{\pi}}

Where A1 A_1 and A2 A_2 are the areas of Circle A and Circle B, respectively.

If d<r1+r2 d < r_1 + r_2 , then the circles overlap.

Let's calculate:

For Circle A:

  • Center: (3,4) (3, 4)
  • Radius: r1=18ππ=18=32 r_1 = \sqrt{\frac{18 \pi}{\pi}} = \sqrt{18} = 3\sqrt{2}

For Circle B:

  • Center: (8,1) (8, 1)
  • Radius: r2=40ππ=40=210 r_2 = \sqrt{\frac{40 \pi}{\pi}} = \sqrt{40} = 2\sqrt{10}

Now, calculate the distance between the centers:

d=(83)2+(14)2=52+(3)2=25+9=34d = \sqrt{(8 - 3)^2 + (1 - 4)^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}

So, d=34 d = \sqrt{34} .

Now, compare d d with r1+r2 r_1 + r_2 :

32+21010.73\sqrt{2} + 2\sqrt{10} \approx 10.7

Since 345.83<10.7 \sqrt{34} \approx 5.83 < 10.7 , the circles overlap.

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Answer from HIX Tutor

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.

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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