Monday, December 19, 2011

Conics - Circles on a coordinate plane

Hey everyone

Today we started our conics unit, and the first topic is circles on a coordinate plane. There are a few important equations to remember:

Standard form of a circle: x2 + y2 = r2 AND (x - h)2 + (y - k)2 = r2

General form of a circle: Ax2 + By2 + Cx + Dx + E = 0

Area of a circle: π r2

Circumference of a circle: 2 π r

Distance between two points: D = √(x2 - x1)2 + (y2 – y1)2

Midpoint between two points: MP = (x2 + x1) , (y2 + y1)
2 2


The most basic operation with a circle is to find the center and radius.
Ex.:
(x-3)2 + (y+4)2 = 48

we find the center by using the h and k values, and reading them as opposites.

Center @ (3, -4)

the radius is the square root of the number after the equal sign.

radius is √48


The second operation we worked on is finding the center, radius, domain, and range of a circle in general form.
Ex.
x2 + y2 – 6x – 10y – 2 = 0

We reorganize the equation so that all like terms are together.

x2 - 6x + y2 – 10y = 2

Then we complete the squares by dividing the C and D values by 2 and squaring them. We add these numbers to both sides.

x2 - 6x + 9 + y2 – 10y + 25 = 2 + 9 + 25

Then we solve each set for one binomial that is squared. Now it is solved.

(x - 3)2 + (y - 5)2 = 36

Next, we find the canter and radius the same way as before.

Center @ (3, 5)
Radius = 6

After sketching the circle, the points are at (-3, 5), (3, -1), (3, 11), and (9, 5). This helps us find the domain and range.

Domain = [-3, 9]
Range = [-1, 11]


The next operation is very similar, but has coefficients.
Ex.
16x2 + 16y2 – 8x – 143 = 0

16x2 – 8x + 16y2 = 143

16(x2 – 1/2x) + 16y2 = 143

We do the same thing with dividing the second number by two, and squaring it.

16(x2 – 1/2x + 1/16) + 16y2 = 143

Then we divide all the numbers in the equation by 16.

16(x2 – 1/2x + 1/16) + 16y2 = 144
16

We then continue to solve as normal.

(x – 1/4)2 + y2 = 9

center @ (¼, 0)
radius = 3


The last operation is using the distance and midpoint formulas.
Ex.
End points at (-2, 8) and (3, 4).

D = √(x2 - x1)2 + (y2 – y1)2
D = √(-2-3)2 + (8-4) 2
D = √(-5)2 + (4) 2
D = √25 + 16
D = √41

Radius = √41/2

MP = (x2 + x1) , (y2 + y1)
2 2

MP = (3 + 2) , (4 + 8)
2 2

MP = (½, 12/2)

MP = (½, 6)

Center @ (½, 6)

Then we plug it in to the formula:

(x – 1/2)2 + (y - 6)2 = √41/2

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