Been stuck on this one help would be appreciated

Let the transmitter be located at (0,0)

The equation for its effective coverage area will be a circle  centered at  (0,0)  with a radius  of  27

So....the equation is   

x^2 + y^2  = 27^2

x^2 + y^2  = 729

Let the point 35  miles north of the transmitter be  (0, 35)

Let the point 31  miles east of the trasmitter   be (31,0)

The line connecting these (the line of travel)  will have a slope  of  [ 35 - 0] / [ 0 - 31]  =  -35/31

The equation of this line will be

y  = (-35/31)x  + 35

We could solve this  algebraically, but a graph seems easier to deal with  

See here : https://www.desmos.com/calculator/ksd2p5jdwl

The signal will be received  betwen the points  (8.221 , 25.718)  and (26.523, 5.055)

The distance between these points  is  

√ [ ( 26.523 - 8.221)^2  + ( 25.718 - 5.055)^2 ]  ≈  27.6 miles     (1)

The distance  between  (0, 35)  and (31,0)  =  √ [ ( 31)^2  + ( 35)^2 ]  ≈  46.8  miles  (2)

So....you will receive the signal  about  (1)/ (2)    =  27.6 / 46.8  ≈  59%  of the drive

We need to find the  x intersection points  of

x^2 + y^2   = 729     and

y = (-35/31)x + 35

So....subbing the second equation into the first for y  we have that

x^2  +  [ (-35/31)x + 35]^2  = 729     

x^2  +  1225x^2/961  -2450x/31  + 1225  = 729

[961 + 1225]x^2 / 961 - 2450x/31  + 496  = 0     multiply through by 961

2186  x^2  -  75950x  + 476656  =  0

Using the quadratic formula    we have that 

75950  ±√ [ (75950^2 - 4(2186)(476656) ]

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         2 *  2186

Evaluating this we get that  x =  8.221  and  x = 26.523

Putiing the first value into  the linear equation we have that

y = (-35/31)(8.221) + 35   ≈ 25.718

Putting the second vallue into the linear eqation we have that

y = (-35/31)(26.523) + 35  ≈ 5.055

So....the intersection points of the circle and line of travel  are ( 8.221, 25.718)  and (25.523, 5.055)

And  I just used the distance formula twice to solve the rest of the problem