+0

0
140
5

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\$.

Mar 20, 2020
edited by Guest  Mar 20, 2020

#1
+21931
0

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 ]

Mar 20, 2020
#3
0

Would you mind solving that so I can confirm my calculations? I feel like I got an unrealistic number :

sqrt[(42sqrt3)/5]

Guest Mar 20, 2020
edited by Guest  Mar 20, 2020
#4
0

Nevermind, got it!

Guest Mar 20, 2020
#2
+21931
0

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)

Mar 20, 2020
#5
+21931
0

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

Mar 20, 2020