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Mathematics 11 Online
OpenStudy (anonymous):

An equation of a circle is x^2 + y^2 – 10x + 6y + 18 = 0. Part 1: Show all your work in determining the center and radius of this circle. (3 points) Part 2: In complete sentences, explain the procedure used. (3 points)

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

@nincompoop

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

you need to factor the equation you have into this format: (x + a)^2 + (y + b)^2 = z^2

OpenStudy (anonymous):

I don't understand how to do that.... Can you help me?

OpenStudy (anonymous):

complete the square, do you know that?

OpenStudy (anonymous):

I don't remember all of it...

OpenStudy (anonymous):

group the like terms together and I like the constant on the right of the equals

OpenStudy (anonymous):

for the x portion, take half of the -10x term and square it

OpenStudy (anonymous):

Sure! when you multiply a binomial like (x + a)^2 you get an x^2 term, a 2a*x term and an a^2 term. You need to look for a way to divide the original eq's x^1 term in half

OpenStudy (anonymous):

and do the same for the y variable. Look at (y + b)^2 and try to find the factor b that doubles to get 6y

OpenStudy (anonymous):

so that would be x^2-10x+25, but since you're adding 25 to the left hand side you need to do that to the right as well. then x^2-10x+25 is written as (x-5)^2

OpenStudy (anonymous):

\[x^2-10x+25+y^2+6y=-18+25\]=\[(x-5)^2+y^2+6y=7\] then do that for the y terms

OpenStudy (anonymous):

take half of the 6y term and square it then add that number to the right side as well \[(x-5)^2+(y+3)^2=16\]

OpenStudy (anonymous):

are you following this lol?

OpenStudy (anonymous):

so we add 9 to the right side?

OpenStudy (anonymous):

yeah, that's how we got 16. so now that it is in the general form for a circle, what is the center and radius?

OpenStudy (anonymous):

center is (5,-3) and the radius is 4?

OpenStudy (anonymous):

A+

OpenStudy (anonymous):

yay! Now can you help me put it into sentences? That's the other part....

OpenStudy (anonymous):

well, tell me what we did

OpenStudy (anonymous):

We rewrote the x part of the equation using completing the square. Then, the same with the y part. Then, we got the center from the x and y parts and the radius from the square root of the term to the right of the equal sign

OpenStudy (anonymous):

but I have to make that sound better lol

OpenStudy (anonymous):

you did pretty good though. I would say something like "We completed the square to get it in the form (x-h)^2+(y-k)^2=r^2, which is the general form of a circle where the center is at (h,k) and radius r.......

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