Answer: C
The radius of the inscribed circle of AHJK is equal to the inradius of the triangle, which is denoted as r.
Since P is the incenter of the triangle, the inradius is equal to the distance from P to each of the sides of the triangle. Let's call these distances a, b, and c.
You know that LP = 4x + 10 and MP = 8x - 2. The length of the inradius can be expressed as:
r = (a+b-c)/2
You can now use this equation and the given information to solve for r.
Since LP is the distance from P to side L, we can set a = 4x + 10.
Since MP is the distance from P to side M, we can set b = 8x - 2.
Substituting these values into the equation for r, you get:
r = ((4x + 10) + (8x - 2) - c) / 2
= (12x + 8 - c) / 2
You can simplify this to:
r = 6x + 4 - c/2
Thus, the radius of the inscribed circle of AHJK is equal to 6x + 4 - c/2. To find the exact value of the radius, you need to know the value of c.