Please help me solve this question! Thanks!

Question

A square region ABCD is externally tangent to the circle with equation x²+y²=1 at the point (0,1) on the side CD. Vertices A and B are on the circle with equation x²+y²=4. What is the side length of this square?

Answer 1

Hi Caroline,

 

Below is the link to the revised diagram as per Selam's corrective comment below. 

  

https://dl.dropbox.com/s/yse20uk9l0yph40/Square_Betweeen_Concentric_Circles.png?raw=1

 

The coordinates of the four vertices of the square are given in reference to the point of tangency (0,1) of line CD to the inner circle:

    1.  Points A and C are x units to the left of the y-axis (x=0) and Points B and D are the same x units to the right of the y-axis (x=0).

    2.  Points C and D lie on the line y=1 and points A and B are both (y-1) units above the line y=1.

    3.  Therefore, the lengths of the horizontal sides of the green square (AB and CD) are each the difference in the x coordinates (2x) and the lengths of the vertical sides of the green square (AC and BD) are each the difference in the y coordinates (y-1).

 

Substituting one variable in terms of the other by equating the lengths of the sides of the square:

    x in terms of y:

        2x = y -1

          x = (y - 1) / 2

 

Then plug this "y" expression for "x" into the equation of the outer circle and solve for the y coordinate to determine the length of a side of the square by subtracting 1 from it as per length of vertical sides (AC and BD) both being y - 1 (see Selam's solution).

 

God bless, Jordan.

Answer 2

As you draw the square ABCD, you will notice half of it will be in Quadrant I and the per half in Quadrant II, with side AB parallel the x-axis passing through (0,1).

Given C and D are on circle with equation (x² +y²=4), let's assign the points: C (x,y) and D (-x,y).

Let the length of the square be represented by s, to write s in terms of x and y,

s=2x

s=(y-1)

therefore, 2x=(y-1)

      x =(y-1)/2

Now, determine the values of x and y in order to find s, the length of the side of the square;

1. Plug in  x =(y-1)/2 in the (x² +y² = 4).                                                        so,           [(y-1)/2]² + y² =4
2. Solve for y, (y-1)² +4y² =16,                                                          then,          y² -2y +1 + 4y² =16

.                          5y²-2y -15=0 (solve using quadratic formula)

y=[2 ± √(316) ]/10

y≈1.98

The solution is,

       s =(1.98-1)

          = 0.98

 

 

The length of the side of the square ABCD is approximately 0.98 units.