Mike@Magoosh wrote: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) 7/3

(B) 4

(C) 5

(D) sqrt(7)

(E) sqrt(11)

For a set of Coordinate Geometry practice questions, including the OA and an OE with a diagram for this problem, see:
https://magoosh.com/gmat/2013/gmat-quan ... questions/

Mike

Mathsbuddy wrote:Using the Equation of a Circle is certainly the right way to go.
However for speed, you could try this:
With a range from y = -1 to y = 7, it is easy to recognise that the radius is 4.
As the centre of the circle is not lying on the x-axis, it means that it must intersect the x-axis at a value less than 4. Ihis eliminates answers B and C straight away.
In fact, a very rough sketch shows that the x-intercepts have a magnitude just less than 3.
Only answer D complies with this.

GMATGuruNY wrote:No special knowledge of circles is required here.
A quick drawing yields the following:

Since (0 -1,) and (0, 7) are both on the circle, AD is a diameter.
Thus, center O is located at (0, 3) -- halfway between (0, -1) and (0, 7) -- implying that r=4.
Since (0, p) is on the circle, OC is also a radius, implying that OC=4.
Since OB=3, and OB² + BC² = OC², we get:
3³ + BC² = 4�
BC² = 7
BC = √7.
Thus, p = √7.

The correct answer is D.

GMATinsight wrote:

Mathsbuddy wrote:Using the Equation of a Circle is certainly the right way to go.
However for speed, you could try this:
With a range from y = -1 to y = 7, it is easy to recognise that the radius is 4.
As the centre of the circle is not lying on the x-axis, it means that it must intersect the x-axis at a value less than 4. Ihis eliminates answers B and C straight away.
In fact, a very rough sketch shows that the x-intercepts have a magnitude just less than 3.
Only answer D complies with this.

Sqrt(7) = 2.64
7/3 = 2.33

Very precise approximation indeed...