GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In the xy-coordinate system, a circle with radius 30^(1/2)
[#permalink]
11 Dec 2018, 14:20
11 Dec 2018, 14:20
Expert Reply
3
Bookmarks
00:00
Question Stats:
70% (02:23) correct
29% (02:24) wrong based on 47 sessions
Hide Show timer Statistics
In the xy-coordinate system, a circle with radius \(\sqrt {30}\) and center (2,1) intersects the x-axis at (k,0). One possible value of k is.
A. \(2 + \sqrt{26}\)
B. \(2 +\sqrt{29}\)
C. \(2 + \sqrt{31}\)
D. \(2 + \sqrt{34}\)
E. \(2 + \sqrt{35}\)
A. \(2 + \sqrt{26}\)
B. \(2 +\sqrt{29}\)
C. \(2 + \sqrt{31}\)
D. \(2 + \sqrt{34}\)
E. \(2 + \sqrt{35}\)
Re: In the xy-coordinate system, a circle with radius 30^(1/2)
[#permalink]
12 Dec 2018, 09:26
12 Dec 2018, 09:26
1
1
Bookmarks
Carcass wrote:
In the xy-coordinate system, a circle with radius \(\sqrt {30}\) and center (2,1) intersects the x-axis at (k,0). One possible value of k is.
A. \(2 + \sqrt{26}\)
B. \(2 +\sqrt{29}\)
C. \(2 + \sqrt{31}\)
D. \(2 + \sqrt{34}\)
E. \(2 + \sqrt{35}\)
A. \(2 + \sqrt{26}\)
B. \(2 +\sqrt{29}\)
C. \(2 + \sqrt{31}\)
D. \(2 + \sqrt{34}\)
E. \(2 + \sqrt{35}\)
Explanation::
The distance (X) between two points (a,b) and (c,d) in a co ordinate plane is given by:
\(X = (\sqrt{(c - a)^2 + (d - b)^2}\)
Here,
\(X =\sqrt{30}\)
c = k and d = 0
a = 2 and b = 1
Therefore,
\(\sqrt{30} = \sqrt{(k - 2)^2 + (0 - 1)^2}\)
\(30 = (k - 2)^2 + 1\)
\((k - 2)^2 = 29\)
Solving for k;
\(K = 2 + \sqrt{29}\)
In the xy-coordinate system, a circle with radius 30^(1/2)
[#permalink]
15 Aug 2022, 19:25
15 Aug 2022, 19:25
1
More intuitively:
We have a right triangle with height 1 and hypotenuse sqrt(30).
Well, Pythagoras can give us a hand:
c^2 = (sqrt(30))^2 - 1^2
c^2 = 29
c = sqrt(29)
c is the length of the basis of the triangle. We just need to remember that we are departing from the point (2,0). So, just add 2 to get the coordinate.
2 + sqrt(29).
We have a right triangle with height 1 and hypotenuse sqrt(30).
Well, Pythagoras can give us a hand:
c^2 = (sqrt(30))^2 - 1^2
c^2 = 29
c = sqrt(29)
c is the length of the basis of the triangle. We just need to remember that we are departing from the point (2,0). So, just add 2 to get the coordinate.
2 + sqrt(29).
