1) Complete the square for the general analytic equation:

Ax

^{2}+ Ay^{2}+ 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}+ Ay^{2}+ 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 = r

^{2}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 y^{2}+ 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)

Radius:

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)

Radius:

r

^{2}= (2^{2}- 4(-8)) / 4 A^{2}
So:

r

^{2}= 36/ 4 = 9
So; r =

**Ö**9 = 3The 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:

x

^{2}+ y^{2}+ 2x = 8_{}

^{}

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