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If the circle in the figure above is centered at the origin of the coo

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If the circle in the figure above is centered at the origin of the coo [#permalink]

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If the circle in the figure above is centered at the origin of the coordinate axes, which of the following coordinates represents a point that lies on the circle?

A. (3,4)
B. (5,5)
C. (1,9)
D. (8,6)
E. (6,6)

[Reveal] Spoiler:
Attachment:
T6192.png
T6192.png [ 6.7 KiB | Viewed 1817 times ]
[Reveal] Spoiler: OA

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Re: If the circle in the figure above is centered at the origin of the coo [#permalink]

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Bunuel wrote:
Image
If the circle in the figure above is centered at the origin of the coordinate axes, which of the following coordinates represents a point that lies on the circle?

A. (3,4)
B. (5,5)
C. (1,9)
D. (8,6)
E. (6,6)

[Reveal] Spoiler:
Attachment:
T6192.png


Radius of the circle shown is 10.

So, equation of the circle can be written as \(x^2 + y^2 = 10^2\)

Now, only point that satisfies this equation is (8,6). Hence, D
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Re: If the circle in the figure above is centered at the origin of the coo [#permalink]

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New post 07 Jan 2017, 12:41
Distance from 0 to the the y intercept of the circle is 10. So pick the coordinates whos squares add upto a 100. Since root of 100 = distance from origin = 10.
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Re: If the circle in the figure above is centered at the origin of the coo [#permalink]

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On questions that involve right triangles, you can often save considerable time by looking for the classic Pythagorean triplets (the 3:4:5 classic triangle and its multiples, the 5:12:13 classic triangle and its multiples, etc.)

Here, a radius drawn from the origin to a point on the circle will be the hypotenuse of a right triangle whose legs are given by the coordinates of the point on the circle.

That hypotenuse/radius will have a length 10 (from the picture). A 3:4:5 triangle is the most likely triangle since it is the only very common classic right triangle with a multiple of 5 for the hypotenuse (it's also the most common right triangle on the gmat). If we're going to have a 10 for the longest side, we want the 6:8:10 version of the classic 3:4:5. Answer D satisfies.
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Re: If the circle in the figure above is centered at the origin of the coo [#permalink]

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New post 23 Jan 2017, 03:42
Bunuel wrote:
Image
If the circle in the figure above is centered at the origin of the coordinate axes, which of the following coordinates represents a point that lies on the circle?

A. (3,4)
B. (5,5)
C. (1,9)
D. (8,6)
E. (6,6)

[Reveal] Spoiler:
Attachment:
T6192.png


From the given figure we have radius = 10
so the equation of the given circle is \(x^2 + y^2 = 10^2\)
only (8,6) satisfies the equation

Hence option D is correct
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Re: If the circle in the figure above is centered at the origin of the coo [#permalink]

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New post 06 Apr 2018, 07:34
Can someone explain this to me? I understand the radius is 10 and that you could make a right triangle between X and Y, but I don't understand how that helps you find a point on the circle. Wouldn't all points on the arc be outside of the triangle?
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If the circle in the figure above is centered at the origin of the coo [#permalink]

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Bunuel wrote:
If the circle in the figure above is centered at the origin of the coordinate axes, which of the following coordinates represents a point that lies on the circle?

A. (3,4)
B. (5,5)
C. (1,9)
D. (8,6)
E. (6,6)
lostnumber wrote:
Can someone explain this to me? I understand the radius is 10 and that you could make a right triangle between X and Y, but I don't understand how that helps you find a point on the circle. Wouldn't all points on the arc be outside of the triangle?

lostnumber , if by "arc" you mean the circumference line, no.
A point lies ON a circle if it lies ON the circumference

Points on a circle

• any point that lies on the circle will
satisfy the equation for the circle (below)

• any point that lies on the circle will
create a right triangle whose hypotenuse = radius

Basic equation of a circle
To determine whether a point lies on a circle,
plug the x- and y-coordinates into the equation

As abhimahna and 0akshay0 mention explicitly,
the basic equation of a circle is

\(x^2 + y^2 = r^2\)

\(x\) and \(y\) are coordinates
of a point on the circle
\(r\) = radius

Basic equation of circle \(<->\) Pythagorean theorem

Pythagorean theorem, where c = hypotenuse:
\(a^2 + b^2 = c^2\)
Basic equation of circle with center (0,0), r = radius:
\(x^2 + y^2 = r^2\)

x replaces a, y replaces b
and r, radius, replaces c, hypotenuse

Any point (x,y) on this circle therefore
satisfies the equation for THIS circle:
\(x^2 + y^2 = 10^2\)
\(x^2 + y^2 = 100\)

Scan the answers.
You're looking for x- and y-values
whose squares will sum to 100.

Plug in answers?
You can plug in until you find the right pair.

Try A. (3,4)

\(3^2 + 4^2 = r^2\)
\(r^2 = 25\)
\(r = 5\)
r must = 10

You can keep plugging in
until you get the correct x- and y- values

See the RHS of diagram.*
All points except D lie INSIDE the circle
Their radii lengths are shorter than 10

OR notice: 6-8-10 = 3-4-5 right triangle

Answer D is (8,6)
Radius = 10
8: 6: 10 = 4x-3x-5x

D's x- and y-coordinates yield lengths
that are a "Pythagorean triplet" triangle

(8, 6) makes a right triangle. See diagram.
Its hypotenuse is the radius.

Check: do D's coordinates (8,6) satisfy equation?

Answer D: (8, 6)
\(x^2 + y^2 = r^2\)
\(8^2 + 6^2 = r^2\)
\(64 + 36 = r^2\)
\(r^2 = 100\)
\(r = 10\)

That's a match.

Answer D

Hope that helps

* Look at the green triangle on the left.
The green point is (-6, 8). It, too, satisfies the equation

\((-6)^2 + 8^2 = r^2\)
\((36 + 64) = 100 = r^2\)
\(r^2 = 100\)
\(r = 10\)

For more on the equation of a circle, see Bunuel , Circle on a plane , and another site HERE

Attachment:
T6192ed2.png
T6192ed2.png [ 35.27 KiB | Viewed 230 times ]

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Re: If the circle in the figure above is centered at the origin of the coo [#permalink]

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New post 09 Apr 2018, 06:59
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Thanks so much for the detailed explanation Generis! It turns out this is simply a math formula and concept that I didn't know, so I'll have to add this to my long list of quant topics to study. But your answer was very detailed and I've bookmarked for future reference. I know what I need to study now! Thanks again
Re: If the circle in the figure above is centered at the origin of the coo   [#permalink] 09 Apr 2018, 06:59
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