Question

On a single set of axes, sketch a picture of the graphs of the following four equations: y = −x+ √ 2, y = −x− √ 2, y = x+ √ 2, and y = x − √ 2. These equations determine lines, which in turn bound a diamond shaped region in the plane.

(a) Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once.

(b) Find the intersection points between the unit circle and each of the four lines.

(c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help.

Answer 1

( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 ), ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )

y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 , y + 1 = ( x + 2 ) - √ 2

y + 1 = ( x + 2 ) + √ 2 , ( x + 2 )^2 + ( y + 1)^2 = 1

Explanation:

Given:

- Four functions to construct a diamond:

y = −x+ √ 2, y = −x− √ 2, y = x+ √ 2, and y = x − √ 2.

Find:

a)Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once.

b)Find the intersection points between the unit circle and each of the four lines.

(c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help.

Solution:

- For first part see the attachment.

- The equation of the unit circle is given as follows:

x^2 + y^2 = 1

- To determine points of intersection we have to solve each given function of y with unit circle equation for set of points of intersection:

For: y = −x+ √ 2 , x - √ 2

And: x^2 + y^2 = 1

x^2 + (+/- * (x - √ 2))^2 = 1

x^2 + (x - √ 2)^2 = 1

2x^2 -2√ 2*x + 2 = 1

2x^2 -2√ 2*x + 1 = 0

2[ x^2 - √ 2] + 1 = 0

Complete sqr: (1 - 1/√ 2)^2 = 0

x = 1/√ 2 , x = 1/√ 2

y = -1/√ 2 + √ 2 = 1/√ 2

y = 1/√ 2 - √ 2 = - 1/√ 2

Points are: ( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 )

- Using vertical symmetry of unit circle we can also evaluate other intersection points by intuition:

x = - 1/√ 2

y = 1/√ 2 , -1/√ 2

Points are: ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )

- To determine the function for the rhombus region that would be tangential to unit circle with center at ( - 2 , - 1 ):

- To shift our unit circle from origin to ( - 2 , - 1 ) i.e two units left and 1 unit down.

- For shifts we use the following substitutions:

x = x + 2 ....... 2 units of left shift

y = y + 1 .......... 1 unit of down shift

- Now substitute the above shifting expression in all for functions we have:

y = −x+ √ 2 -----> y + 1 = - ( x + 2 ) + √ 2

y = −x− √ 2 -----> y + 1 = - ( x + 2 ) - √ 2

y = x- √ 2 -------> y + 1 = ( x + 2 ) - √ 2

y = x+ √ 2 ------> y + 1 = ( x + 2 ) + √ 2

x^2 + y^2 = 1 -----> ( x + 2 )^2 + ( y + 1)^2 = 1

- The following diamond shape graph would have the 4 functions as:

y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 , y + 1 = ( x + 2 ) - √ 2

y + 1 = ( x + 2 ) + √ 2 , ( x + 2 )^2 + ( y + 1)^2 = 1

- See attachment for the new sketch.