1) The line y = {3x + 15}{4} intersects the circle x^2 + y^2 = 36 at A and B. Find the length of chord AB.
2) The real numbers x and y satisfy x^2 + y^2 = 4x + 4y. Find the largest possible value of x.
3) Let A = (4,-1), B = (6,2), and C = (-1,2). There exists a point $X$ and a constant $k$ such that for any point P, PA^2 + PB^2 + PC^2 = 3PX^2 + k. Find the constant $k$.
+23246 y = (3x + 15)/4 x2 + y2 = 36
Substituting: x2 + [ (3x + 15)/4 ]2 = 36
x2 + (9x2 + 45x + 225) / 16 = 36
16x2 + 9x2 + 45x + 225 = 576
25x2 + 45x - 351 = 0
Using the quadratic formula: x1 = [-9 + 12sqrt(3) ] / 5 or x2 = [-9 - 12sqrt(3) ] / 5
Substituting x1 into y = (3x + 15)/4 results in y1 = [12 + 9sqrt(3) ] / 5
Substituting x2 into y = (3x + 15)/4 results in y2 = [12 - 9sqrt(3) ] / 5
A = ( [-9 + 12sqrt(3) ] / 5 , [12 + 9sqrt(3) ] / 5 )
B = ( [-9 - 12sqrt(3) ] / 5 , [12 - 9sqrt(3) ] / 5 )
Now, to find the length of AB, place the values of (x1, y1) and (x2, y2) into the distance formula:
Distance = sqrt[ (x2 - x1)2 + (y2 - y1)2 ]
+23246 2) x2 + y2 = 4x + 4y
x2 + y2 - 4x - 4y = 0
(x2 - 4x ) + (y2 - 4y ) = 0
(x2 - 4x + 4) + (y2 - 4y + 4) = 8 (complete the square)
(x - 2)2 + (y - 2)2 = ( 2sqrt(2) }2
The largest value of x will occur when the y-term becomes zero; so, let y = 2:
(x - 2)2 + (2- 2)2 = ( 2sqrt(2) }2
(x - 2)2 + (0)2 = ( 2sqrt(2) }2
(x - 2)2 = ( 2sqrt(2) }2
x - 2 = 2sqrt(2)
x = 2 + 2sqrt(2)
+23246 3) Let the coordinates of point P be (x, y).
Using the distance formula:
PA = sqrt[ (x - 4)2 + (y - -1)2 ] ---> PA = sqrt[ (x - 4)2 + (y + 1)2 ] ---> PA2 = (x - 4)2 + (y + 1)2
PB = sqrt[ (x - 6)2 + (y - 2)2 ] ---> PB = sqrt[ (x - 6)2 + (y +- 2)2 ] ---> PB2 = (x - 6)2 + (y - 2)2
PC = sqrt[ (x - -1)2 + (y - 2)2 ] ---> PC = sqrt[ (x + 1)2 + (y - 2)2 ] ---> PC2 = (x + 1)2 + (y - 2)2
Combining these:
PA2 + PB2 + PC2 = (x - 4)2 + (y + 1)2 + (x - 6)2 + (y - 2)2 + (x + 1)2 + (y - 2)2
= (x2 - 8x + 16) + (y2 + 2y + 1) + (x2 - 12x + 36) + (y2 - 4y + 4) + (x2 + 2x + 1) + (y2 - 4y + 4)
= 3x2 - 18x + 3y2 - 6y + 62
= (3x2 - 18x ) + 3y2 - 6y ) + 62
= 3(x2 - 6x ) + 3(y2 - 2y ) + 62
= 3(x2 - 6x + 9) + 3(y2 - 2y + 1) + 32
= 3(x - 3)2 + 3(y - 1)2 + 32
Let the coordinates of X be (a,b) ---> PX = sqrt[ (x - a)2 + (y- b)2 ] ---> PX2 = (x - a)2 + (y- b)2
3PX2 + k = 3[ (x - a)2 + (y- b)2 ] + k = 3(x - a)2 + 3(y - b)2 + k
Setting these two equal to each other: 3(x - 3)2 + 3(y - 1)2 + 32 = 3(x - a)2 + 3(y - b)2 + k
tells me that a = 3, b = 1, and k = 32
