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Given the two lines y = 2x + 5 and y = 2x - 10, what is the

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Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 02 Sep 2012, 19:05
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Aug 2014, 20:31, edited 1 time in total.
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Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 02 Sep 2012, 19:17
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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


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_jee-Strai ... 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] New post 29 Sep 2012, 12:23
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_jee-Strai ... 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 all,

Pls explain the question above !!!
What is a and b in he above formula ?
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Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 29 Sep 2012, 13:23
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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


[Reveal] Spoiler:
(45/4)pi



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 x-intercept and the y-intercept 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 x-intercept and the y-intercept 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] New post 09 Oct 2012, 07:09
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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 bi-sector 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 co-ord 1 is (-2,1)

Next
y=2x-10
y=-1/2x
x=4
Subbing 4 into y=2x-10=-2
Co-ord 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] New post 15 Oct 2012, 09:44
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] New post 15 Oct 2012, 10:03
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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] New post 16 Oct 2012, 05:58
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] New post 11 Nov 2012, 14:42
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_jee-Strai ... 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

y-2x-5 = 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] New post 31 Jul 2013, 03:13
@bunuel can u pls help n explain . i cant understand either
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Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 02 Aug 2013, 06:30
to clarify what nkdotgupta wrote

write the equation in the form of ax+by+c

Eq 1. y = 2x + 5 can be written as 2x-y+5 = 0 which makes a = 2 b = -1 and c1 = 5

Eq 2. y = 2x - 10 can be written as 2x-y-10 = 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] New post 22 Aug 2014, 14:30
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Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 23 Aug 2014, 10:20
Point (0,5) satisfies equation y=2x+5
Point (5,0) satisfies equation y=2x-10
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] New post 23 Aug 2014, 11:10
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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=2x-10
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
Untitled.png [ 12.54 KiB | Viewed 743 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] New post 23 Aug 2014, 12:00
Bunuel wrote:
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=2x-10
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
So, this line segment cannot be the diameter of the circle.

Hope it's clear.


Hi Bunuel,

I got your point. Can you please suggest how to derive the length of perpendicular using graphical approach.
I am not able to get it.
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Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 24 Aug 2014, 03:50
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] New post 24 Aug 2014, 06:21
Graphical approach:

Straight lines with equation f(x) = 2x + 5 and g(x) = 2x - 10 are parallel; slope of each line equal to 2.

Co-ordinate of y-intercepts 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 y-intercept of g(x):h(x) = -(1/2)x-10

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}x-10 orx =-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

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Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink] New post 28 Aug 2014, 05:59
Temurkhon wrote:
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


yes, I did the same thing. 5 min. Too bad :|
Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the   [#permalink] 28 Aug 2014, 05:59
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