Circles and lines Pre Calculus

Question

CIRCLES AND LINES PRE CALCULUS

I need help answering these two questions for a final exam. Please help 

1.) Let A(1,3) and B(4,−2).  Find all points P on the y-axis such that AP⊥BP (that is ,∠APB= 90◦).

2.) Let A(2,4) and B(8,3).  Find the point P on thex-axis such that AP+PB is minimum.

Answer 1

1

A is at (1, 3)

B is at (4, -2)

Let P be on the y-axis

Then, P is at (0,y)

P- A = (0,y) - (1,3) = (-1, y -3)

P-B =(0, y) -(4,-2) = (-4, y + 2)

Then, for AP and BP to be orthogonal,

(-1,y - 3) . (-4, y +2) =0

-1* -4 + (y-3)(y + 2) =0

4 + y² -y -6 =0

y² -y - 2 = 0

(y - 2)(y +1) = 0

y = 2

y = -1

P is (0,-1) or (0, 2)

2

There is a trick here to the solution. Note that the 2 points currently are both above the y-axis (y is 4 and 3), so no line can be drawn through P that connects the two points.

Flip the y coordinates on the second point, call this B', find the line AB', and the point where the line crosses the x-axis is the solution.

Why this is the solution: BP and B'P have the same length and the point P that minimizes AP + PB' is the one that has P on the line AB'.

Then, A is (2, 4), B is (8, 3), and B' is (8, -3)

Solve for the line that goes through AB'

The slope is (-3 -4)/(8 - 2) =-7/6

Using point A to solve for the y-intercept,

4 = 2(-7/6) + b

4 = -7/3 + b

b = 4 + 7/3 or 19/3

The equation of the line is y = -7/6x + 19/3

Now, we solve for the x-intercept

0 = -7/6x +19/3

7/6x = 19/3

x = 38/7

The point P is (38/7, 0) or (5 3/7, 0)

Answer 2

Let A and B be two points on opposite ends of a circle. The line AB is now the diameter of the circle and any point P on that circle is such that AP and BP form a 90° angle. Try finding the equation of that circle.