# Circle A has a center at #(-1 ,2 )# and a radius of #2 #. Circle B has a center at #(5 ,-4 )# and a radius of #1 #. Do the circles overlap? If not what is the smallest distance between them?

no overlap , ≈ 5.485

What we have to do here is compare the distance (d ) between the centres with the sum of the radii.

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

radius of A + radius of B = 2 + 1 = 3

Since sum of radii < d , then no overlap

smallest distance = 8.485 - 3 = 5.485 graph{(y^2-4y+x^2+2x+1)(y^2+8y+x^2-10x+40)=0 [-10, 10, -5, 5]}

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The circles do not overlap. The distance between the centers of the circles can be calculated using the distance formula: Distance = √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the centers of the circles.

For Circle A: Center at (-1, 2) For Circle B: Center at (5, -4)

Distance = √((5 - (-1))^2 + (-4 - 2)^2) = √((5 + 1)^2 + (-4 - 2)^2) = √((6)^2 + (-6)^2) = √(36 + 36) = √72 ≈ 8.49

Since the sum of the radii of the two circles (2 + 1 = 3) is less than the distance between their centers (8.49), the circles do not overlap. The smallest distance between them is the distance between their centers minus the sum of their radii, which is approximately 8.49 - 3 = 5.49 units.

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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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