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Magoosh GMAT Instructor
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The center of circle Q is on the yaxis, and the circle passes through
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26 Nov 2014, 11:48
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 yaxis, and the circle passes through points (0, 7) and (0, –1). Circle Q intersects the positive xaxis at (p, 0). What is the value of p?
(A) \(\frac{7}{3}\)
(B) \(4\)
(C) \(5\)
(D) \(\sqrt{7}\)
(E) \(\sqrt{11}\)For a set of Coordinate Geometry practice questions with solutions, visit this page: https://magoosh.com/gmat/2013/gmatquan ... questions/Mike
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Re: The center of circle Q is on the yaxis, and the circle passes through
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10 Jun 2015, 18:30
The center of circle Q is on the yaxis, 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 xaxis at (p, 0). we can just use the equation for distance between two points. In this case, (0,3) and (p,0).
4=\(sqrt((03)^2 + (p0)^2)\) p=\(sqrt(7)\)
Answer: (D) \(sqrt(7)\)



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Re: The center of circle Q is on the yaxis, and the circle passes through
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02 Dec 2015, 19:15
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 ycoordinate 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 yaxis, and the circle passes through
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13 Sep 2019, 14:37
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 (71)/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((x2x1)^2 + (y2y1)^2) 16 = (p0)^2 + (30)^2 16 = p^2 + 9 p^2= 7 p = root 7
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Re: The center of circle Q is on the yaxis, and the circle passes through
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