To avoid doing people's homework, I'll work a problem exactly like yours,
but with different numbers.
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Find the coordinates of P if P is 17 units from L(-3,-6) and 10 units from M(-1,1)

It appears that there are two solutions.

OR

We use the distance formula, setting PL = 17 and PM = 10

d = 

 = 17
 = 10

Squaring both sides of both equations:

(x+3)² + (y+6)² = 17²
(x+1)² + (y-1)² = 10²

x²+6x+9 + y²+12y+36 = 289
x²+2x+1 +   y²-2y+1 = 100

Simplifying

x²+y²+6x+12y = 244
x²+y²+2x- 2y =  98

Subtracting those two equations:

      4x+14y = 146

Divide that equation through by 2

      2x+ 7y = 73
          2x = 73-7y
           x = 

Substitute in

x²+y²+2x- 2y =  98
 + y² + 2 - 2y = 98

 + y² + 73 - 7y - 2y = 98

 + y² - 9y  = 25

Clear the fraction by multiplying through by 4

5329 - 1022y + 49y² + 4y² - 36y = 100

            53y² - 1058y + 5229 = 0

That factors as

             (53y - 581)(y - 9) = 0

           53y - 581 = 0;          y - 9 = 0
                 53y = 581;            y = 9
                   y = 

Substitute each in 2x + 7y = 73

             2x + 7y = 73
2x + 7() = 73
         106x + 4067 = 3869
                106x = -198
                   x = 
                   x = 

So one solution is the point P(, )

[That's approximately the point P(-1.97,10.96)

The other solution is

             2x + 7y = 73
           2x + 7(9) = 73
             2x + 63 = 73
                  2x = 10
                   x = 5

So the other solution is P(5,9)

Edwin