A radar buoy detects any boats within aradius of 12 miles. A rowboat starts at a location 18 miles south and 10 miles east of the radar buoy. The rowboat travels at aconstant speed of 15 mph. The tugboattravels on a straight line toward the northermostpoint of the radar region. When the tugboat is directlyeastof the buoy, it turns and travelsdue northuntil it exits the radar region.

How long (in hours) is the tugboat in the radar region?

This is what I did:

I represented it by a circle, letting the radar be the center (0,0)

Here are my steps:

1) x^{2}+ y^{2}= 12^{2}

2) x^{2}+ y^{2}= 144

The initial position P(10,-18) is heading to (0,12)

So I find the linear equation of the line:

3) Slope = (12-(-18))/(0-(10)) = -3 , with a

Y-intercept is 12

So first path: y = -3x + 12

Where it crosses

The circle:

5) x^{2}+ (-3x+12)^{2}= 144

6) x^{2}+ 9x^{2}+ 72x + 144 = 144

7) 10x^{2}+ 72x = 0

8) x(10x + 72) = 0

9) x = 0 or x = 7.2

So now I know x=0 is my point

Our point (0,12)

If x = 7.2

10) y = 3(7.2) + 12 = 33.6

So the intersection of his path with the radar region is (7.2, 33.6)

I crosses the x-axis at x = -4 , (let y=0 in y = 3x+12 )

So first leg is from (7.2 , 33.6) to (-4,0) , then it heads north

so let x = -4 in the circle equation:

16 + y^{2 }= 144

y = √128 , which is the distance it travels northbound

within the region.

Finally find the distance between (-7.2, -9.6) and

(-4,0) , add on √128 for the total distance.

Since Time = distance/rate

Distance between (-7.2, -9.6) and (-4,0) would be

d=√(-4-(-7.2))^{2}+(0-33.6)) = √((3.2)^{2}+(-33.6)^{2})

d=1139.2

11) Time = 1139.2/15 =75.94 hours

It seems WAY too much! (I tried getting help from some other people, and they say the answer is 5.5 hours in radar region, though I don't know how they got it so I don't know if they got it right).

Is my first leg correct?

Please help

Thanks!