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Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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. \(45\pi\)
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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.



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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!
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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 \(\pi\cdot9\cdot{5}/4=(45/4)\pi\).
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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.



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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 \(\sqrt{9\cdot{5}}=\sqrt{45}\).
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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!
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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.



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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..



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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.



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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 7986 times ]
So, this line segment cannot be the diameter of the circle. Hope it's clear.
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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)
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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...



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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?



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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.



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



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Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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



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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the [#permalink]
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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?
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Re: Given the two lines y = 2x + 5 and y = 2x  10, what is the
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