This is a blog for my second semester pre-calculus 30s students. Have a blast!

Monday, March 20, 2006

HI:)every body I’m the scribe for today
Today we start a new unit (the forth unit), and it’s called:

--> consider a fixed point. A circle is:

A circle is the set of all points (x, y) that are equidistant from a fixed point called the centre of the circle. The distance, r, between the centre of the circle and any point (x, y) on the circle is the radius.

--> What if the center of the circle is (h,K)
(x-h)^2+(y-k)^2=r^2 this is the Std. form.

-->expanded form:
(x-h)^2+(y-k)^2=r^2 ... where d,e,f E real #.


For the circle:

x^2+y^2-8x+12y-35=0 {Rewrite with like variables together and constant on the right.}

i) Find center:

1. (x^2-8x)+(y^2+12y)=35 {Group the like variables and begin to complete the square of each group.}

2. (x^2-8x+16)+(y^2+12y+36)=35+16+36 {Take half the middle term and square it to complete the square. Add to both sides.}


center= (4,-6)

ii) r=squ. root of 87

Ex.: Wirte the eqn. of a circle with center (-1,-2) and passing through (3,4)
first we have to find the r using the distance formula:
then place it back at the eqn.

Ex: Find the eqn of a circle with diameter AB, with A(1,2) & (-3,-6)

first we have to find the mid. point , in order to find the r, because half of diameter is =to r. then use mid. point. with one of the point from que. and solve.


find the eqn. of the circle with center (0,4) and circumference of 8pi.

area of cirucm=2pi(r)





there for x^2+(y-4)^2=16


the home work for tomorrow is ex#21:)

and the scribe for tomorrow is jojo


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