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A circle with center (h, k), in which k = 2h, lies tangent to the y-ax

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A circle with center (h, k), in which k = 2h, lies tangent to the y-ax  [#permalink]

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New post 10 Feb 2019, 03:30
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A
B
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Question Stats:

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A circle with center (h, k), in which k = 2h, lies tangent to the y-axis in in the (x, y) coordinate system. What is the distance from the origin to the center of the circle, in terms of k ?

A. \((k\sqrt{3)}/2\)
B. \((k\sqrt{5)}/2\)
C. k
D. k \(\sqrt{3}\)
E. k \(\sqrt{5}\)

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Re: A circle with center (h, k), in which k = 2h, lies tangent to the y-ax  [#permalink]

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New post 10 Feb 2019, 04:04
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SajjadAhmad wrote:
A circle with center (h, k), in which k = 2h, lies tangent to the y-axis in in the (x, y) coordinate system. What is the distance from the origin to the center of the circle, in terms of k ?

A. \((k\sqrt{3)}/2\)
B. \((k\sqrt{5)}/2\)
C. k
D. k \(\sqrt{3}\)
E. k \(\sqrt{5}\)



Let us draw the sketch first as per the info..

If we can draw the sketch, we are almost 90% there..Tangent means that a point of the circle lies on the y-axis.
This point on y-axis, the center of the circle and the origin will make a RIGHT angled triangle.

Distance = \(\sqrt{k^2+h^2}=\sqrt{k^2+(k/2)^2}=\sqrt{5k^2/4}=\sqrt{5}*\frac{k}{2}\)

B
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A circle with center (h, k), in which k = 2h, lies tangent to the y-ax  [#permalink]

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New post Updated on: 13 Feb 2019, 22:24
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Solution



Given
    • A circle is present in the (x, y) coordinate system, whose center is (h, k)
    • The circle is tangent to the y-axis
    • Also, h = 2k, or k = h/2

To Find
    • The distance from the origin to the center of the circle, in terms of k

Approach & Working
Image
As we can see in the figure, angle OAC = 90° (as the circle is tangent to the y-axis)

Therefore, as per Pythagoras Theorem,
    • \(OC^2 = OA^2 + AC^2 = k^2 + \frac{k^2}{4} = \frac{5k^2}{4}\)
    Or, OC = k\(\sqrt{5}/2\)

Hence, the correct answer is option B.

Image

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Originally posted by EgmatQuantExpert on 10 Feb 2019, 04:09.
Last edited by EgmatQuantExpert on 13 Feb 2019, 22:24, edited 1 time in total.
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Re: A circle with center (h, k), in which k = 2h, lies tangent to the y-ax  [#permalink]

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New post 10 Feb 2019, 04:38
1
SajjadAhmad wrote:
A circle with center (h, k), in which k = 2h, lies tangent to the y-axis in in the (x, y) coordinate system. What is the distance from the origin to the center of the circle, in terms of k ?

A. \((k\sqrt{3)}/2\)
B. \((k\sqrt{5)}/2\)
C. k
D. k \(\sqrt{3}\)
E. k \(\sqrt{5}\)



k=2h
center ( h,k)
use the distance formula Sqrt( y2-y1)^2+(x2-x1)^2
and determine the distance from origin (0,0) to center of circle
= > k * sqrt5/2
IMO B
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Re: A circle with center (h, k), in which k = 2h, lies tangent to the y-ax  [#permalink]

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New post 21 Mar 2019, 05:43
1) Draw a diagram to get an idea of how to solve
2) Plug in numbers for h and k, h=2 k=4
3) Solve using pythagorean theorem, 2²+4² = hypotenuse² --> hyp = 2√5
4) Plug into answers, if k=4 then it's clearly B.
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Re: A circle with center (h, k), in which k = 2h, lies tangent to the y-ax   [#permalink] 21 Mar 2019, 05:43
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