Logic to solve this

I am new to GMAT but here are my two cents.

I arrive at the answer 45pi/4. Not sure where I am going wrong.

The slopes of both the lines are 2 which means that these are parallel lines. Now the largest circle that is tangent to both the lines would have the diameter as the perpendicular distance between the lines. So the sum is all about finding the distance between two parallel lines.

The formula for distance between two parallel lines is as follows:
Line 1: y = mx + c1
Line 2: y = mc + c2
Distance = absolute value of (c1 - c2) / sqr root ( m2 + 1 ) ..........here m2 = m square

Diameter becomes 15/sqr root(5) -> Radius = 15/2root(5)

Area = 45pi/4

[email protected] 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!

Answer is 45pi

y = 2x + 5 ____________(1)
y = 2x - 10 ____________(2)

Both the lines are of the form y = mx + c.
Clearly, slopes m are the same.
Since the slopes are the same the two lines are parallel.
This means the greatest diameter the circle can have with the conditions given is equal to the distance between the 2 lines.

My solution:
To find the distance, I used a line perpendicular to both these lines: y = (-1/2)x ____________(3)

Solving (1) and (3):
(-1/2)x = 2x + 5
(-5/2)x = 5
x = -2
y = 1

Solving (2) and (3):
(-1/2)x = 2x - 10
(5/2)x = 10
x = 4
y = -2

So diameter = distance between (-2,1) and (4,-2) = sqrt(((-2)-4)^2 + (1-(-2))^2)
= sqrt(36 + 9)
= sqrt(45)

Radius R = sqrt(45)/2
Area = pi * R^2
= 45pi/4