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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. \(45\pi\)
A. \(\frac{45}{4}*\pi\)
B. \(2\sqrt{45}\pi\)
C. \(27\pi\)
D. \(27\sqrt{2}\pi\)
E. \(45\pi\)
Updated on: 26 Jun 2024, 04:21
Originally posted by BrushMyQuant on 02 Sep 2012, 19:17.
Last edited by BrushMyQuant on 26 Jun 2024, 04:21, edited 11 times in total.
Last edited by BrushMyQuant on 26 Jun 2024, 04:21, edited 11 times in total.
↧↧↧ Detailed Video Explanation to the Problem ↧↧↧
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 = \(\frac{|C_1–C_2|}{\sqrt{}(A^2+B^2)}\)
Here \(C_1\) = 5, \(C_2\)=-10
A = 2
B = -1
So, d = \(\frac{|5–(-10)|}{\sqrt{}(2^2+(-1)^2)}\) = \(\frac{|15|}{\sqrt{}5}\) = \(\frac{3 * 5}{\sqrt{}5}\)
=> d = \(3\sqrt{}5\)
=> Area = \(\frac{(\pi * d^2)}{4}\)
= \(\pi * \frac{(3\sqrt{}5)^2}{ 4}\)
= \(\frac{45 \pi}{4}\)
So, Answer will be A
Hope it helps!
Watch the following video to Learn Basics of Circles
_________________
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 = \(\frac{|C_1–C_2|}{\sqrt{}(A^2+B^2)}\)
Here \(C_1\) = 5, \(C_2\)=-10
A = 2
B = -1
So, d = \(\frac{|5–(-10)|}{\sqrt{}(2^2+(-1)^2)}\) = \(\frac{|15|}{\sqrt{}5}\) = \(\frac{3 * 5}{\sqrt{}5}\)
=> d = \(3\sqrt{}5\)
=> Area = \(\frac{(\pi * d^2)}{4}\)
= \(\pi * \frac{(3\sqrt{}5)^2}{ 4}\)
= \(\frac{45 \pi}{4}\)
So, Answer will be A
Hope it helps!
Watch the following video to Learn Basics of Circles
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dvinoth86
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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General Discussion
Posts: 28
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
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
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.
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.
IanSolo
\(\frac{15}{\sqrt{5}}=\frac{15\sqrt{5}}{\sqrt{5}\sqrt{5}}=\frac{15\sqrt{5}}{5}=3\sqrt{5}\) which is the same as \(\sqrt{9\cdot{5}}=\sqrt{45}\).
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PhD in Applied Mathematics
Love GMAT Quant questions and running.
Love GMAT Quant questions and running.
IanSolo
d = 15/sqrt5 = 3sqrt5
area = (pie * d^2)/4
= (pie * (3sqrt5)^2)/4
= (pie * 45/4)
hope it helps!
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How to Solve :
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
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nktdotgupta
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.
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..
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..
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.
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.
desaichinmay22
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 22082 times ]
Hope it's clear.
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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
|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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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, \(2x+5=\frac{-1}{2}x-10\) 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)
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, \(2x+5=\frac{-1}{2}x-10\) 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
Distance-Between-Parallel-Lines.png [ 24.44 KiB | Viewed 21844 times ]
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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...
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...
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 => |C1-C2|/√(A^2+B^2)
so the distance => 15/√5 => 3√5 => Radius = 3√5/2 => area =45/4 * pie => A is correct
Now the diameter of the circle that is tangent to both => distance between the 2 || lines => |C1-C2|/√(A^2+B^2)
so the distance => 15/√5 => 3√5 => Radius = 3√5/2 => area =45/4 * pie => A is correct
EvaJager
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?
BrushMyQuant
Can you explain what is A & B. I checked the link but could not get it.
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A and B are the constant values of the variables x and y respectively.
dvinoth86
There are two ways I can think of to solve this problem.
1)
The distance between two parallel lines is given by the equation |B-C| / (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 y-intecept, we get Y=-.5X + 5. This line crosses the first line at the y-intecept.
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
BrushMyQuant
Could you please tell how to find A=2 and B=-1?
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