## Saturday, January 21, 2017

### Solutions to Analytic Geometry Problems (1)

The problems again - with solutions to follow:

1)  Complete the square for the general analytic equation:

Ax 2    +  Ay2   + Dx  + Ey  + F   = 0

To show how the radius r is derived, as well as the coordinates for the center of a circle not necessarily at (0,0)

Solution:

Write:  (x 2   + Dx / A) +  (y 2   + Ey/ A) =    - F/A

[x 2   + Dx / A  +   (D/ 2A) 2  ] +   [y 2   + Ey / A  +   (E/ 2A)2 ]

= - F/A + (D 2   +    E  2  /  4 A 2

SO:

(x  +   D/ 2A)2   +  (y  +   E/ 2A)2       =   D 2  +   E 2      - 4AF) /  4 A 2

Then:   r 2     = (D 2   +    E 2      - 4AF) /  4 A 2

Looking at the two factors we see the center coordinates must be:

h =  (-D/ 2A)   and k =  (-E/ 2A)

2) Check the equation:  x 2    +    y 2  =  4

to see that the general form also applies here.

Solution:

Compare to general form:

Ax 2    +  Ay2   + Dx  + Ey  + F   = 0   (A not equal 0)

Then: A = 1 , D = 0, E = 0, F  =   -4

The center is at:

h = (-D/ 2A)   = 0   and k =  (- E/ 2A) = 0  So:   C(0, 0)

Since   4  =  r2  then r  is 2, then we see radius is 2.

3) Find the coordinates of the center of each of the following circles and the radius r, also sketch the circle:

a)  x 2    +    y 2  - 2 y   =  3

We have: A = 1, D = 0,  E = -2  and F = -3

Then the center is at:

(h, k) =    (-D/ 2A),   (-E/ 2A)  =   (0,  1)

r 2     = (D 2   +    E 2      - 4AF) /  4 A 2

So:

r 2     = (0 2   +    (-2)  2      - 4(-3_) /  4 (1) 2

r 2     =   16/ 4   = 4  so radius  r  = 2

b) 2x 2    +  2  y2   + x  + y = 0

Here: A = 2,   D = 1,  E = 1,  F = 0

Center is at:

(h, k) =   (-D/ 2A),   (-E/ 2A)  =   (- 1/4,  -1/4)

r 2      = (D 2   +    E 2      - 4AF) /  4 A 2

So:

r 2     = (1 2   +    (1)  2      - 4(0) /  4 (1) 2

=   2/ 4 =  1/2

So:  r =  1/ Ö 2

c) x 2    +    y 2  + 2x  = 8

We have: A = 1,  D= 2, E = 0, F = -8

Center is at:

(h, k) =   (-D/ 2A),   (-E/ 2A)  =   (- 2/ 2,  0 ) = (-1, 0)

r 2      = (2 2     - 4(-8)) /  4 A 2

So:

r 2     =    36/ 4   =  9

So; r =  Ö 9   =    3

The circle is the same as shown below for (4)

4) Obtain the equation for the circle shown below: By inspection, center is at (h, k) =   (-1, 0)

Also, radius r = 3 units

We now write the most general form using coordinates (h, k) for the center:

(x - h) 2    + (y - k ) 2   =  r 2

Then:

(x - (-1))2    + (y - 0 ) 2   =  3 2

The equation for the circle is then:

(x + 1)2    + y 2   =   9

Or:

x 2    +  2x  + 1  +    y 2   = 9

Finally:

2    +    y 2  +  2x   =  8