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# In the figure to the right, the circle is centered at the origin and p

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Status: Preparing GMAT
Joined: 02 Nov 2016
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In the figure to the right, the circle is centered at the origin and p  [#permalink]

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02 Jun 2017, 10:30
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Difficulty:

25% (medium)

Question Stats:

76% (01:26) correct 24% (01:11) wrong based on 104 sessions

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In the figure to the right, the circle is centered at the origin and passes through point P. Which of the following points does it also pass through?

(A) (3, 3)
(B) (-2 $$\sqrt{2}$$, -1)
(C) (2, 6)
(D) (-3 $$\sqrt{3}$$, $$\sqrt{3}$$)
(E) (–3, 4)

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Re: In the figure to the right, the circle is centered at the origin and p  [#permalink]

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02 Jun 2017, 11:04
1
Since point P(0,-3) passes through the circle,
the distance between the point P and origin is $$\sqrt{(0 - 0)^2 + (0-(-3))^2}$$ or 3
Formula used :
Distance between 2 points (x1,y1) and (x2,y2) is $$\sqrt{(x1-x2)^2 + (y1-y2)^2}$$

Any other point on the circle will also have a distance of 3 units from the origin.
On evaluating the answer choices, only Option B has the distance of 3
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Re: In the figure to the right, the circle is centered at the origin and p  [#permalink]

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02 Jun 2017, 14:49
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1
In the figure to the right, the circle is centered at the origin and passes through point P. Which of the following points does it also pass through?

(A) (3, 3)
(B) (-2 $$\sqrt{2}$$, -1)
(C) (2, 6)
(D) (-3 $$\sqrt{3}$$, $$\sqrt{3}$$)
(E) (–3, 4)

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I used the equation of the circle. For a circle to pass through a point, that point's coordinates must satisfy the circle's equation.

General formula: $$(x - h)^2 + (y - k)^2 = r^2$$

With center at origin, substitute (0,0) for x and y; and from point (0, -3), substitute those numbers respectively for h and k to find r.

(0 - 0)$$^2$$ + (0 - -3)$$^2$$ = r$$^2$$

= $$0 + 9 = r^2$$

$$r = 3$$

So the circle's equation is $$x^2 + y^2 = 3^2$$, or

$$x^2 + y^2 = 9$$

Answers - A, C, D, and E can all be ruled out quickly. If, as is the case with A and E, a point has one coordinate whose absolute value is 3, to satisfy this equation, the other coordinate would have to be 0, because both $$(-3)^2$$ and $$3^2$$ $$= 9$$:

$$(-3)^2 + y^2 = 9$$

$$9 + y^2 = 9$$

$$y^2 = 0$$

y must equal = 0. Reject A and E.

And if, as is the case with C and D, one coordinate's absolute value is greater than 3 (6, and -3 $$\sqrt{3}$$, respectively), its squared value will certainly be greater than 9, let alone when added to the square of the other coordinate's absolute value.

So (B), with coordinates $$(-2\sqrt{2}, -1)$$:

$$(-2\sqrt{2})^2$$ + $$(-1)^2$$ =

$$8 + 1 = 9.$$ Correct.

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In the figure to the right, the circle is centered at the origin and p  [#permalink]

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13 Apr 2018, 09:26
2
I solved by determining that points (0, -3), (0,3), (3,0) must be the other intersections with the X/Y axis because we know the radius is 3. Once I knew these points I could rule out A and E, because I already knew the correct points on the circle for 3, Y and -3, Y. Then I realized that since the radius is the distance from the center to the edge, there couldn't be a value that exceeded 3 or -3, otherwise it would be outside the circle. That let me rule out C and D

So there is a non-math formula way to determine the answer
In the figure to the right, the circle is centered at the origin and p &nbs [#permalink] 13 Apr 2018, 09:26
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