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The center of circle Q is on the y-axis, and the circle pass

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The center of circle Q is on the y-axis, and the circle pass

by Mike@Magoosh » Wed Nov 26, 2014 10:50 am
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/

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by GMATinsight » Wed Nov 26, 2014 10:43 pm
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 :-)
Circle passes through (0, 7) and (0, -1) and has centre on Y-axis which means that distance between (0, 7) and (0, -1) should be diameter
i.e. Diameter = 8
i.e. Radius = 4
and Co-ordinate of Centre = (0, 3)

Since general equation of circle with centre (h, k) is
(X-h)^2 +(Y-k)^2 = r^2
Therefore, Equation of circle becomes X^2 +(Y-3)^2 = 4^2
i.e. X^2 +(Y-3)^2 = 16

for circle to intersect with X axis, y co-ordinate should be zero therefore, equation of circle becomes
X^2 +(0-3)^2 = 4^2
i.e. X = Sqrt(16-9)
i.e. p = Sqrt(7)

Answer: Option D

P.S. This is not a GMAT Like question as GMAT talk about the equation of circle with centre at origin and doesn't expect students to know that general equation of circle with centre (h, k) is
(X-h)^2 +(Y-k)^2 = r^2
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by Mathsbuddy » Thu Nov 27, 2014 2:05 am
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.
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by GMATinsight » Thu Nov 27, 2014 3:56 am
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... :P ;)
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by GMATGuruNY » Thu Nov 27, 2014 3:56 am
No special knowledge of circles is required here.
A quick drawing yields the following:
Image
Since (0 -1) and (0, 7) are both on the circle, and the center of the circle is on the y-axis, 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:
3² + BC² = 4²
BC² = 7
BC = √7.
Thus, p = √7.

The correct answer is D.
Last edited by GMATGuruNY on Thu Nov 27, 2014 4:25 am, edited 3 times in total.
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by GMATinsight » Thu Nov 27, 2014 3:59 am
GMATGuruNY wrote:No special knowledge of circles is required here.
A quick drawing yields the following:
Image
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.
Nice one Mitch. Just a few typo errors in powers on the step [3³ + BC² = 4�]
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by GMATGuruNY » Thu Nov 27, 2014 4:09 am
Thanks, Bhoopendra. I've corrected the typos.
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by Mathsbuddy » Thu Nov 27, 2014 8:03 am
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... :P ;)
True! I forgot to dismiss this on the grounds that 7/3 = Sqrt(49/9), which containing a fraction, which could not exist in this case as all our rectilinear given dimensions are integer. (To be honest I did overlook this. Thanks)
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