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In the coordinate plane, a diameter of a circle has the end points (−3

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In the coordinate plane, a diameter of a circle has the end points (−3  [#permalink]

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07 Jul 2017, 00:48
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In the coordinate plane, a diameter of a circle has the end points (−3, −6) and (5, 0). What is the area of the circle?

(A) $$5\pi$$

(B) $$10\sqrt{2}*\pi$$

(C) $$25\pi$$

(D) $$50\pi$$

(E) $$100\pi$$

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In the coordinate plane, a diameter of a circle has the end points (−3  [#permalink]

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07 Jul 2017, 03:25
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Bunuel wrote:
In the coordinate plane, a diameter of a circle has the end points (−3, −6) and (5, 0). What is the area of the circle?

(A) $$5\pi$$

(B) $$10\sqrt{2}*\pi$$

(C) $$25\pi$$

(D) $$50\pi$$

(E) $$100\pi$$

It should be C.

In coordinate plane, Modulus of difference of X and Y coordinates of two respective points can serve as base and height of the triangle whose hypotenuse would be the Diameter of the circle in this case.

Base = 5 - (-3) = 8
Height = 0 - (-6) = 6

From Pythagoras theorem,

Diameter^2 = $$8^2 + 6^2$$

Diameter = 10

Radius = 5

Area = $$π * 5^2 = 25π$$
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Re: In the coordinate plane, a diameter of a circle has the end points (−3  [#permalink]

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07 Jul 2017, 08:17
1
Bunuel wrote:
In the coordinate plane, a diameter of a circle has the end points (−3, −6) and (5, 0). What is the area of the circle?

(A) $$5\pi$$

(B) $$10\sqrt{2}*\pi$$

(C) $$25\pi$$

(D) $$50\pi$$

(E) $$100\pi$$

Distance between end points of diameter = $$\sqrt{(8)^2 + (6)^2}$$
= 10
Radius = 5
Area = $$25\pi$$

Answer C
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Re: In the coordinate plane, a diameter of a circle has the end points (−3  [#permalink]

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14 Jul 2017, 09:12
End points of the circle is given on graph (-3,-6) & (5,0)
Calculate the distance formula - Distance =10
Radius=5

Area of Circle = pi*r^2 = 25pi
Answer C
Re: In the coordinate plane, a diameter of a circle has the end points (−3   [#permalink] 14 Jul 2017, 09:12
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In the coordinate plane, a diameter of a circle has the end points (−3

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