Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Oct 2011
Posts: 114
Location: India

Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
Updated on: 22 Aug 2014, 21:31
Question Stats:
64% (02:14) correct 36% (02:17) wrong based on 525 sessions
HideShow timer Statistics
Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines A. \(\frac{45}{4}*\pi\) B. \(2\sqrt{45}\pi\) C. \(27\pi\) D. \(27\sqrt{2}\pi\) E. \(45\pi\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Encourage me by pressing the KUDOS if you find my post to be helpful. Help me win "The One Thing You Wish You Knew  GMAT Club Contest"http://gmatclub.com/forum/theonethingyouwishyouknewgmatclubcontest140358.html#p1130989
Originally posted by dvinoth86 on 02 Sep 2012, 20:05.
Last edited by Bunuel on 22 Aug 2014, 21:31, edited 1 time in total.
Edited the question.




Director
Status: Tutor  BrushMyQuant
Joined: 05 Apr 2011
Posts: 613
Location: India
Concentration: Finance, Marketing
GPA: 3
WE: Information Technology (Computer Software)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
02 Sep 2012, 20:17
Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines Diameter of the circle will be equal to the distance between these two parallel tangents. Distance between parallel lines is given by the formula d = C1–C2/√(A^2+B^2) (refer: http://www.askiitians.com/iit_jeeStrai ... llel_lines) Here C1 = 5, C2=10 A = 2 B = 1 So d = 5–(10)/√(2^2+(1)^2) => d = 3√5 So area = (pie d^2)/4 = pie * (3√5(^2) )/ 4 = 45 pie /4 Hope it helps!
_________________
Ankit
Check my Tutoring Site > Brush My Quant
GMAT Quant Tutor How to start GMAT preparations? How to Improve Quant Score? Gmatclub Topic Tags Check out my GMAT debrief
How to Solve : Statistics  Reflection of a line  Remainder Problems  Inequalities




Director
Joined: 22 Mar 2011
Posts: 604
WE: Science (Education)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
29 Sep 2012, 14:23
dvinoth86 wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines Even if you don't know the formula for the distance between two parallel lines, you can figure it out. The two lines are parallel, having the same slope 2. A circle tangent to both lines will have the diameter equal to the distance between the two lines. To find this distance, take the perpendicular from the origin to each of the lines. The origin and the xintercept and the yintercept of the line \(y = 2x  10\) form a right triangle with legs 5 and 10 and hypotenuse \(5\sqrt{5}\). Therefore, the height corresponding to the hypotenuse is \(10\cdot{5}/5\sqrt{5}=2\sqrt{5}\) (use the formula height = leg*leg/hypotenuse). The distance between the origin and the line \(y = 2x + 5\) is half of the distance between the origin and the line \(y = 2x  10\), because the xintercept and the yintercept of the line \(y = 2x + 5\) are 5/2 and 5, which with the origin, form a right triangle similar to the right triangle discussed above. So, the diameter of the circle is \(3\sqrt{5}\) and the area of the circle is \(\pi\cdot9\cdot{5}/4=(45/4)\pi\).
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Intern
Joined: 12 Jun 2012
Posts: 37

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
09 Oct 2012, 08:09
My answer by using first principles and not the A and B formula Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines The first thing to use is the perpendicular bisector between the two lines. If y=2x is the gradient of the lines, then y= 1/2x is the gradient of the bisector. Step 1, solve the two simultaneous equations in order to find the vectors of the line you seek. y=2x+5 y= 1/2x x = 2 Subbing 2 into y=2x+5=1 So coord 1 is (2,1) Next y=2x10 y=1/2x x=4 Subbing 4 into y=2x10=2 Coord to is (4,2) Now you have a pythagorus equation Triangle base = 6 Triangle height = 3 Hypotenuse (the diameter of the circle) ^2 = 36 + 9 Hyp= diameter = sqrt(45) Radius = Sqrt(45)/2 Area = Pi x r^2 =45/4 x pi
_________________
If you find my post helpful, please GIVE ME SOME KUDOS!



Intern
Joined: 09 Oct 2012
Posts: 32

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
15 Oct 2012, 10:44
nktdotgupta I have tried to solve with your way, but when I try to calculate d I receive another result: \(d = \frac{15}{\sqrt{5}}\) Then I divide d with 2 for find r and eventually I elevate r^2 and moltiplicate with pie, but the result is different. I'm not able to find the error.



Director
Joined: 22 Mar 2011
Posts: 604
WE: Science (Education)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
15 Oct 2012, 11:03
IanSolo wrote: nktdotgupta I have tried to solve with your way, but when I try to calculate d I receive another result: \(d = \frac{15}{\sqrt{5}}\) Then I divide d with 2 for find r and eventually I elevate r^2 and moltiplicate with pie, but the result is different. I'm not able to find the error. \(\frac{15}{\sqrt{5}}=\frac{15\sqrt{5}}{\sqrt{5}\sqrt{5}}=\frac{15\sqrt{5}}{5}=3\sqrt{5}\) which is the same as \(\sqrt{9\cdot{5}}=\sqrt{45}\).
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Director
Status: Tutor  BrushMyQuant
Joined: 05 Apr 2011
Posts: 613
Location: India
Concentration: Finance, Marketing
GPA: 3
WE: Information Technology (Computer Software)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
16 Oct 2012, 06:58
IanSolo wrote: nktdotgupta I have tried to solve with your way, but when I try to calculate d I receive another result: \(d = \frac{15}{\sqrt{5}}\) Then I divide d with 2 for find r and eventually I elevate r^2 and moltiplicate with pie, but the result is different. I'm not able to find the error. d = 15/sqrt5 = 3sqrt5 area = (pie * d^2)/4 = (pie * (3sqrt5)^2)/4 = (pie * 45/4) hope it helps!
_________________
Ankit
Check my Tutoring Site > Brush My Quant
GMAT Quant Tutor How to start GMAT preparations? How to Improve Quant Score? Gmatclub Topic Tags Check out my GMAT debrief
How to Solve : Statistics  Reflection of a line  Remainder Problems  Inequalities



VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1004
Location: India
GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
11 Nov 2012, 15:42
nktdotgupta wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines Diameter of the circle will be equal to the distance between these two parallel tangents. Distance between parallel lines is given by the formula d = C1–C2/√(A^2+B^2) (refer: http://www.askiitians.com/iit_jeeStrai ... llel_lines) Here C1 = 5, C2=10 A = 2 B = 1 So d = 5–(10)/√(2^2+(1)^2) => d = 3√5 So area = (pie d^2)/4 = pie * (3√5(^2) )/ 4 = 45 pie /4 Hope it helps! Hi a very fundamental problem with the explanation. Had the two eqns of the form y2x5 = 0 and y  2x +10 =0 which is same to the above eqn, should the eqn be converted to y = mx + c form for getting c1 and c2 i.e the eqn Ax + By + C = 0 needs to be converted to form y = mx+ c or not.



Intern
Joined: 17 May 2013
Posts: 41
GMAT Date: 10232013

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
02 Aug 2013, 07:30
to clarify what nkdotgupta wrote
write the equation in the form of ax+by+c
Eq 1. y = 2x + 5 can be written as 2xy+5 = 0 which makes a = 2 b = 1 and c1 = 5
Eq 2. y = 2x  10 can be written as 2xy10 = 0 which makes a = 2 b = 1 and c2 = 10
hope this helps..



Manager
Joined: 21 Sep 2012
Posts: 218
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
23 Aug 2014, 11:20
Point (0,5) satisfies equation y=2x+5 Point (5,0) satisfies equation y=2x10 Take (0,0) to form a right triangle with points (0,5) and (5,0) Height of triangle= distance between (0,0) and (0,5) = 5 Base of triangle=distance between (0,0) and (5,0) = 5 Therefore hypotenuse =5\sqrt{2} That is the diameter of the circle. Area of circle= pi*r^2 =pi*(5\sqrt{2}/2)^2 =pi*50/4
Can anyone explain what am I missing in this approach.



Math Expert
Joined: 02 Sep 2009
Posts: 49303

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
23 Aug 2014, 12:10
desaichinmay22 wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines
Point (0,5) satisfies equation y=2x+5 Point (5,0) satisfies equation y=2x10 Take (0,0) to form a right triangle with points (0,5) and (5,0) Height of triangle= distance between (0,0) and (0,5) = 5 Base of triangle=distance between (0,0) and (5,0) = 5 Therefore hypotenuse =5\sqrt{2} That is the diameter of the circle. Area of circle= pi*r^2 =pi*(5\sqrt{2}/2)^2 =pi*50/4
Can anyone explain what am I missing in this approach. The point is that a circle tangent to both those lines must have the diameter equal to the distance between the two lines. The distance between two parallel lines is the length of perpendicular from one to another, while line segment joining (5,0) and (0,5) is NOT perpendicular to the lines: Attachment:
Untitled.png [ 12.54 KiB  Viewed 8435 times ]
So, this line segment cannot be the diameter of the circle. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 23 Jan 2013
Posts: 590

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
24 Aug 2014, 04:50
I used the following formula to find distance between lines: ax+by+c/sqrt(a^2+b^2), got 15/sqrt5, halved and got 15/2*(sqrt5), squared it to get 45/4pi
But it took more than 5 min. to solve



Manager
Joined: 04 Oct 2013
Posts: 155
Location: India
GMAT Date: 05232015
GPA: 3.45

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
24 Aug 2014, 07:21
Graphical approach:Straight lines with equation f(x) = 2x + 5 and g(x) = 2x  10 are parallel; slope of each line equal to 2. Coordinate of yintercepts of lines f(x) and g(x) are (0,5) and (0,10) Equation of the line perpendicular to f(x) and g(x) and passing through yintercept of g(x):h(x) = (1/2)x10 Point of intersection between h(x) and f(x) : (6,7) [ Point of intersection satisfies both line equations f(x) and h(x). Thus, \(2x+5=\frac{1}{2}x10\) or\(x =6\) and \(y=7\)] Distance between parallel lines = Distance between (6,7) and (0,10) = \(\sqrt{( 10+7)^2+(0+6)^2}\)\(=3\sqrt{5}\) Largest circle that can be inscribed between lines f(x) and g(x) must have the diameter equal to the distance between these two lines = \(3\sqrt{5}\) Area of the circle = \(\pi*(3\sqrt{5}/2)^2= \frac{45}{4}*\pi\) Answer: (A)
Attachments
DistanceBetweenParallelLines.png [ 24.44 KiB  Viewed 8288 times ]



Board of Directors
Joined: 17 Jul 2014
Posts: 2683
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
26 Mar 2016, 18:35
wow..this problem is a nightmare..i guessed it..but tried to solve it for more than half an hour... I picked A, because it is written in X/4 pi. I knew that one point from one line to the other point on the other line would form the diameter. to find the radius, we need to divide it by 2. the area would be (d/2)^2 or d^2 / 4 I did not know how to solve, so picked A, as it had /4...



Current Student
Joined: 12 Aug 2015
Posts: 2648

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
04 Apr 2016, 14:32
Here is my approach => The two lines have equal slopes => must be parallel. Now the diameter of the circle that is tangent to both => distance between the 2  lines => C1C2/√(A^2+B^2) so the distance => 15/√5 => 3√5 => Radius = 3√5/2 => area =45/4 * pie => A is correct
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Intern
Joined: 31 Jul 2017
Posts: 7

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
23 Sep 2017, 21:13
EvaJager wrote: dvinoth86 wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines Even if you don't know the formula for the distance between two parallel lines, you can figure it out. The two lines are parallel, having the same slope 2. A circle tangent to both lines will have the diameter equal to the distance between the two lines. To find this distance, take the perpendicular from the origin to each of the lines. The origin and the xintercept and the yintercept of the line \(y = 2x  10\) form a right triangle with legs 5 and 10 and hypotenuse \(5\sqrt{5}\). Therefore, the height corresponding to the hypotenuse is \(10\cdot{5}/5\sqrt{5}=2\sqrt{5}\) (use the formula height = leg*leg/hypotenuse). The distance between the origin and the line \(y = 2x + 5\) is half of the distance between the origin and the line \(y = 2x  10\), because the xintercept and the yintercept of the line \(y = 2x + 5\) are 5/2 and 5, which with the origin, form a right triangle similar to the right triangle discussed above. So, the diameter of the circle is \(3\sqrt{5}\) and the area of the circle is \(\pi\cdot9\cdot{5}/4=(45/4)\pi\). Therefore, the height corresponding to the hypotenuse is \(10\cdot{5}/5\sqrt{5}=2\sqrt{5}\) (use the formula height = leg*leg/hypotenuse). Could you explain this step? What do you mean the height corresponding to the hypotenuse?



Manager
Joined: 04 May 2014
Posts: 161
Location: India
WE: Sales (Mutual Funds and Brokerage)

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
24 Oct 2017, 22:47
BrushMyQuant wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines Diameter of the circle will be equal to the distance between these two parallel tangents. Distance between parallel lines is given by the formula d = C1–C2/√(A^2+B^2) (refer: http://www.askiitians.com/iit_jeeStrai ... llel_lines) Here C1 = 5, C2=10 A = 2 B = 1 So d = 5–(10)/√(2^2+(1)^2) => d = 3√5 So area = (pie d^2)/4 = pie * (3√5(^2) )/ 4 = 45 pie /4 Hope it helps! Can you explain what is A & B. I checked the link but could not get it.



Intern
Joined: 10 Mar 2016
Posts: 37
Location: India
Concentration: Finance, Marketing

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
01 Nov 2017, 12:33
A and B are the constant values of the variables x and y respectively.



Intern
Joined: 03 Aug 2017
Posts: 21
Location: United States (NY)
GPA: 3.76

Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
20 Nov 2017, 23:48
dvinoth86 wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines
A. \(\frac{45}{4}*\pi\)
B. \(2\sqrt{45}\pi\)
C. \(27\pi\)
D. \(27\sqrt{2}\pi\)
E. \(45\pi\) There are two ways I can think of to solve this problem. 1) The distance between two parallel lines is given by the equation BC / (M^2 + 1)^0.5 where line 1 is Y= MX + B and line 2 is Y=MX + C Thus here, we get 5(10) / (2^2 + 1)^.5 or 15/(5^.5) This is the diameter of the circle so, 15/2(5^.5) = radius Pi*R^2 = area, Pi * (15/2(5^.5))^2 = Pi * 225/20 = Pi*45/4 2) To find the distance between the two lines we need to find the perpendicular line. Because the slop is 2, the perpendicular line will have a slope of .5. Starting at the yintecept, we get Y=.5X + 5. This line crosses the first line at the yintecept. Then set this line equal to the second line to get where they cross. .5X + 5 = 2X 10. 15 = 2.5x X = 6 Plug 6 into the equation to find that Y, 2. So we now how the bases of a triangle. from 2 to 5, = 3 and from 0 to 6 = 6. Thus from the Pythagorean theorem (6^2 + 3^2)^.5, we get a hypotenuse of 45^.5 The hypotenuse is the diameter of the circle, so (45^.5)/2 is the radius. Pi * R^2 = Area Pi * ((45^.5)/2)^2 Pi * 45/4



Manager
Status: Turning my handicaps into assets
Joined: 09 Apr 2017
Posts: 125

Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
[#permalink]
Show Tags
07 Jan 2018, 07:39
BrushMyQuant wrote: Given the two lines y = 2x + 5 and y = 2x  10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines Diameter of the circle will be equal to the distance between these two parallel tangents. Distance between parallel lines is given by the formula d = C1–C2/√(A^2+B^2) (refer: http://www.askiitians.com/iit_jeeStrai ... llel_lines) Here C1 = 5, C2=10 A = 2 B = 1 So d = 5–(10)/√(2^2+(1)^2) => d = 3√5 So area = (pie d^2)/4 = pie * (3√5(^2) )/ 4 = 45 pie /4 Hope it helps! Could you please tell how to find A=2 and B=1?
_________________
If time was on my side, I'd still have none to waste......




Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the &nbs
[#permalink]
07 Jan 2018, 07:39






