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 |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