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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
27 Feb 2011, 08:32
2
2
rxs0005 wrote:
In the XY co-ordinate plane , circle C has center at ( 8,0 ) and tangent to the line y = x what is the diameter of the circle
8
4 * root(2)
8 * root(2)
16
16 * root(2)
Refer to the diagram below:
Attachment:
1.PNG [ 15.75 KiB | Viewed 6275 times ]
If a line is tangent to a circle, then a radius drawn to the point of contact is perpendicular to that line.
Now, as the line y=x makes 45 degrees with the axis then we have 45-45-90 right triangle with hypotenuse equal to 8, thus the leg/radius (red segment) equals to \(\frac{8}{\sqrt{2}}\) and the diameter equals to \(2*\frac{8}{\sqrt{2}}=8*\sqrt{2}\).
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
27 Feb 2011, 08:50
1
1
Bunuel - the graph in post above is for a circle at centre (0,8) and not (8,0) as asked in the question
one can use coordinate geometry formula to solve this as well.
Since the line y=x is tangent to the circle with centre (8,0) the length of the perpendicular dropped from point (8,0) to this line would be equal to radius.
Now, length of perpendicular dropped from point (X,Y) to any line ax+by+c=0 is given by the formula Mod ((aX+bY+c)/(a^2+b^2)^0.5))
So, length of perpendicular from point (8,0) is 8/(2)^0.5
and Diameter = 2*radius = 16/(2)^0.5 = 8*(2)^0.5. Answer C
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
27 Feb 2011, 09:01
beyondgmatscore wrote:
Bunuel - the graph in post above is for a circle at centre (0,8) and not (8,0) as asked in the question
Yes, actually it is. Though as the diagram is symmetric to y=x it doesn't matter at all: for any point (8,0), (0,8), (-8,0), (0,-8) the answer would be the same.
Also I wouldn't recommend approach you used as it involves memorizing the formula you don't really need for the GMAT. As you can see one can solve this problem in pretty straightforward way without this formula.
_________________
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
27 Feb 2011, 09:18
Yes - symmetry makes it immaterial in this particular case.
I agree with you about not needing to remember this formula, but if one knows this concept, quite a few questions of similar nature can be done faster - so this might be useful for some people.
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
27 Feb 2011, 09:38
1
beyondgmatscore wrote:
Yes - symmetry makes it immaterial in this particular case.
I agree with you about not needing to remember this formula, but if one knows this concept, quite a few questions of similar nature can be done faster - so this might be useful for some people.
OK then here it is:
DISTANCE BETWEEN THE LINE AND POINT: Line: \(ay+bx+c=0\), point \((x_1,y_1)\)
\(d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}\)
DISTANCE BETWEEN THE LINE AND ORIGIN: As origin is \((0,0)\) -->
\(d=\frac{|c|}{\sqrt{a^2+b^2}}\)
Please note again that it's highly unlikely that you'll need this on the GMAT.
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
10 Sep 2012, 10:19
4
Bunuel wrote:
rxs0005 wrote:
In the XY co-ordinate plane , circle C has center at ( 8,0 ) and tangent to the line y = x what is the diameter of the circle
8
4 * root(2)
8 * root(2)
16
16 * root(2)
Refer to the diagram below:
Attachment:
1.PNG
If a line is tangent to a circle, then a radius drawn to the point of contact is perpendicular to that line.
Now, as the line y=x makes 45 degrees with the axis then we have 45-45-90 right triangle with hypotenuse equal to 8, thus the leg/radius (red segment) equals to \(\frac{8}{\sqrt{2}}\) and the diameter equals to \(2*\frac{8}{\sqrt{2}}=8*\sqrt{2}\).
Answer: C.
Other method to solve such questions. It's slightly lengthy in the beginning but kinda foolproof The point at which line y=x touches the circle is perpendicular to the circle. Slope of line y=x is 1 Thus slope of the line (centre of the circle) touching the point of tangent is -1 & its equation will be (y-0)/(x-8)= -1 y = -x +8..............(1) y = x ...................(2) Point of intersection of line (1) & (2) is x = -x +8 2x = 8 ----> x =4 The other co-ordinate of point of intersection will be y = -x +8---> y = -4+8---->y=4 The co-ordinates of the point of intersection is (4,4) Now the distance between point (4,4) & (8,0) will be the radius of the circle Radius = √((0-4)^2+(8-4)^2 )=4√2 Therefore diameter = 8√2 Answer: C
Hope it will help many others to come.
_________________
If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory
If you have any question regarding my post, kindly pm me or else I won't be able to reply
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
05 Sep 2017, 13:50
Hi bunuel please explain the solution in more steps, it is unclear to me. From where does square root 2 came? please help me to understand this.
_________________
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
05 Sep 2017, 20:13
SandhyAvinash wrote:
Hi bunuel please explain the solution in more steps, it is unclear to me. From where does square root 2 came? please help me to understand this.
Triangle formed is a 45-45-90 right triangle.
• A right triangle where the angles are 45°, 45°, and 90°. This is one of the 'standard' triangles you should be able recognize on sight. A fact you should also commit to memory is: The sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.
For more on this issues check Triangles chapter of Math Book.
_________________
Re: In the XY co-ordinate plane, circle C has center at (8,0) and tangent
[#permalink]
Show Tags
25 Sep 2018, 10:33
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
53% (01:38) correct 
