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In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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11 Oct 2015, 09:25
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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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11 Oct 2015, 10:42
Bunuel wrote: In the coordinate plane, a circle centered on point (3, 4) passes through point (1, 1). What is the area of the circle?
A. 9π B. 18π C. 25π D. 37π E. 41π
Kudos for a correct solution. We can build a triangle with the given points and find the hypotenuse (Radius) using Pythagorean theorem: 4^2+5^2=r^2 > 16+25 so the area is equal to 41Pi Answer (E)
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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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12 Oct 2015, 10:42
Bunuel wrote: In the coordinate plane, a circle centered on point (3, 4) passes through point (1, 1). What is the area of the circle?
A. 9π B. 18π C. 25π D. 37π E. 41π
Kudos for a correct solution. r^2=(31)^2+(41)^2=16+25=41 Area of circle=πr^2=41π Answer : E
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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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12 Oct 2015, 11:08
The circle is with center (3,4) and one of the point as (1,1).Using distance equation we can get the radius : r^2 = sqrt ((31)^2 + (41)^2) or r = 41^1/2
Hence aread = pi (r)^2 = 41 pi



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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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12 Oct 2015, 11:26
From the distance formula: r^2= (31)^2 + (41)^2 r^2= 16+25=41 As we know, Area of a Circle=pi * r^2 Therefore, Area of Circle= 41 pi Hence, the answer is 'E'
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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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18 Oct 2015, 12:04
Bunuel wrote: In the coordinate plane, a circle centered on point (3, 4) passes through point (1, 1). What is the area of the circle?
A. 9π B. 18π C. 25π D. 37π E. 41π
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:As you know, the area of a circle is \(π(r)^2\), meaning that your goal is to find the radius of this circle. To go from the center of the circle (3, 4) to the given point on the circle (1, 1), you move 4 spaces horizontally (from xcoordinate 3 to xcoordinate 1) and 5 spaces vertically (from ycoordinate 4 to ycoordinate 1). That sets up a right triangle in which the hypotenuse is the radius. a^2+b^2=c^2 becomes 4^2+5^2=c^2, so c^2=16+25=41. And here's where a shortcut awaits. Since your job is to take the radius (c in the Pythagorean Theorem) and square it, then multiply by π (π(r)^2) you're actually about done. Since \(r^2=41\), the answer is \(41π\).
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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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14 Jul 2016, 08:32
Bunuel wrote: In the coordinate plane, a circle centered on point (3, 4) passes through point (1, 1). What is the area of the circle?
A. 9π B. 18π C. 25π D. 37π E. 41π
Kudos for a correct solution. Apply distance formula between the xy pair of (3,4) and (1,1) The distance so obtained will be the radius of the circle Once radius is known then \(area = π*r^2\) \(D= \sqrt{(31)^2+(41)^2}\) \(D= \sqrt{(4)^2+(5)^2}\) \(D= \sqrt{16+25}\) \(D= \sqrt{41}\) \(Radius = \sqrt{41}\) \(Radius^2 = 41\) \(Area = Radius^2*π\) \(Area= 41π\) Answer is E
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Re: In the coordinate plane, a circle centered on point (3, 4) passes th [#permalink]
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