To solve the classic problem of finding the area of a square where a point (P) inside has (PA=1), (PB=2), (PC=3), follow these steps:

Step 1: Coordinate Setup & Equations

Let the square have side length (s) (area = (s^2)). Assign coordinates: (B(0,0)), (C(s,0)), (A(0,s)), (P(x,y)).

We get 3 equations:

• (PA^2 = x^2 + (y-s)^2 = 1)
• (PB^2 = x^2 + y^2 = 4)
• (PC^2 = (x-s)^2 + y^2 = 9)

Step 2: Simplify Equations

Subtract (PB^2) from (PA^2):
(x^2 + (y-s)^2 - (x^2 + y^2) = 1-4)
(-2sy + s^2 = -3 \implies sy = \frac{s^2 +3}{2})

Subtract (PB^2) from (PC^2):
((x-s)^2 + y^2 - (x^2 + y^2) = 9-4)
(-2sx + s^2 =5 \implies sx = \frac{s^2 -5}{2})

Step 3: Compute Area (s^2)

Since (x^2 + y^2 =4), multiply by (s^2):
((sx)^2 + (sy)^2 =4s^2)

Substitute (sx) and (sy):
(\left(\frac{s^2 -5}{2}\right)^2 + \left(\frac{s^2 +3}{2}\right)^2 =4s^2)

Let (u=s^2) (area):
(\frac{(u-5)^2 + (u+3)^2}{4}=4u)
((u^2 -10u +25)+(u^2 +6u +9)=16u)
(2u^2 -4u +34=16u)
(u^2 -10u +17=0)

Solve for (u):
(u=\frac{10±\sqrt{100-68}}{2}=5±2\sqrt{2})

Take the positive larger value: (u=5+2\sqrt{2})

Answer: (\boxed{5 + 2\sqrt{2}})

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