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23 Jan 2021, 22:42
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
65% (02:08) correct
35%
(02:13)
wrong
based on 95
sessions
History
Date
Time
Result
Not Attempted Yet
What is the maximum possible distance between the point P (-5, 0) and a point A on the circle \(x^2 + y^2 = 4\) such that line PA intersects the circle at just one point?
A. 4
B. √17
C. √20
D. √21
E. √22
A. 4
B. √17
C. √20
D. √21
E. √22
Source: https://www.GMATinsight.com
24 Jan 2021, 00:39
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GMATinsight
The general equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2, where ( h, k ) is the center and r is the radius.
Now, here the equation is \(x^2+y^2=4=2^2\)
So h=0 and k=0, meaning the Center of the circle is the origin and also the radius =2.
Since the line PA intersects the circle at only one point, the line PA is a tangent to the circle. Thus PAC will form a right angles triangle with side as following
PC is the hypotenuse of the triangle and is equal to 0-(-5)=5
AC is the radius =2
PA : \(PA^2=PC^2-AC^2=5^2-2^2=21......PA=\sqrt{21}\)
So the only possible value of PA is \(\sqrt{21}\), and not the maximum possible value.
30 Jan 2021, 01:34
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GMATinsight
Answer: Option D
Video solution by GMATinsight
Attachment:
Screenshot 2021-01-30 at 1.54.52 PM.png [ 930 KiB | Viewed 3138 times ]
