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)
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 = 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:
x 2 + y 2 + 2x = 8
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