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Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4472
The center of circle Q is on the y-axis, and the circle passes through  [#permalink]

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12 00:00

Difficulty:   55% (hard)

Question Stats: 66% (02:10) correct 34% (02:11) wrong based on 212 sessions

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The center of circle Q is on the y-axis, and the circle passes through points (0, 7) and (0, –1). Circle Q intersects the positive x-axis at (p, 0). What is the value of p?

(A) $$\frac{7}{3}$$

(B) $$4$$

(C) $$5$$

(D) $$\sqrt{7}$$

(E) $$\sqrt{11}$$

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Mike McGarry
Magoosh Test Prep

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Joined: 28 Feb 2014
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Concentration: Strategy, General Management
Re: The center of circle Q is on the y-axis, and the circle passes through  [#permalink]

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1
The center of circle Q is on the y-axis, and the circle passes through points (0, 7) and (0, –1). This tells us the circle's center, which is (0,3) and the radius, which is 4.

Circle Q intersects the positive x-axis at (p, 0). we can just use the equation for distance between two points. In this case, (0,3) and (p,0).

4=$$sqrt((0-3)^2 + (p-0)^2)$$
p=$$sqrt(7)$$

(D) $$sqrt(7)$$
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Re: The center of circle Q is on the y-axis, and the circle passes through  [#permalink]

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agree with D. if center of the circle is on y axis, then it must be the case that r=4 while the center is at 0,3. Now, we can draw a line from 0,3 to the point of intersection of the circle with x axis. the line will have a length of 4, since it is the radius of the circle.
we know 2 sides of a right triangle...we can find the 3rd side: 4^2 = 3^2 +x^2, where x is the length of the leg.
16=9+x^2 -> x=sqrt 7.
since the y-coordinate of the line is 0, then the x coordinate of the intersection of the circle must be sqrt 7.
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Re: The center of circle Q is on the y-axis, and the circle passes through  [#permalink]

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Given that the two points (0,7) and (0,-1) are so far apart but lay on the circle, it is safe to assume the distance that separates them is the diameter and thus the midpoint of these two points will allow us to determine the radius.

(0+0)/2 = 0
(7-1)/2 = 3
midpoint = (0,3)

See plotted

If you plotted correctly you can plainly see the radius is 4.

This radius can be used to identify the point p,0 as 4 is the distance between 0,3 and any point on the circle.

4 = root((x2-x1)^2 + (y2-y1)^2)
16 = (p-0)^2 + (3-0)^2
16 = p^2 + 9
p^2= 7
p = root 7
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+1 Kudos if I have helped you Re: The center of circle Q is on the y-axis, and the circle passes through   [#permalink] 13 Sep 2019, 14:37
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