Let point P be inside rectangle ABCD such that PA = 1, PB = 7 , and PC = 8. Find PD .
+130071 First....let A be located at (0,0)....construct a circle with radius 1 and centered at A
And this circle wil pass through P and have the equation
x^2 + y^2 = 1 → y^2 = 1 - x^2 (1)
And let B be located at (c,0).........construct a circle with radius 7 and centered at B
And this circle will pass through P and have the equation
(x - c)^2 + y^2 = 49 sub (1) in for y^2
(x -c)^2 + [1 - x^2 ] = 49
(x - c)^2 = 48 + x^2 (2)
Next......let C be located at (c,d).........construct a circle with radius 8 and centered at C
And this circle will pass through P and have the equation
(x - c)^2 + (y - d)^2 = 64 sub (2) in for (x - c)^2
[48 + x^2] + (y - d)^2 = 64
(y - d)^2 = 16 - x^2 (3)
Finally......
Let D be located at (0,d).........construct a circle with radius DP and centered at D
And this circle will pass through P and have the equation
x^2 + (y - d)^2 = DP^2 sub (3) in for (y - d)^2
x^2 + [ 16 - x^2] = DP^2 ....... simplify..........
16 = DP^2 take the positive root
4 = DP = PD
