What is the standard form of the equation for this circle?

- anonymous

What is the standard form of the equation for this circle?

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- anonymous

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- anonymous

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@ any one who can help me

- anonymous

The standard form of a circle is \((x-h)^2+(y-k)^2=r^2\) where the center of the circle is \((h, k)\) and \(r\) is the radius. Can you take it from there?

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## More answers

- anonymous

okay so then that would mean it is (x-4)^2 + (y+5)^2 = 5.5^2

- anonymous

right?

- anonymous

Yup. Great job!

- anonymous

thanks!

- anonymous

np :)

- anonymous

@Calcmathlete could you help me with another question?

- anonymous

Sure

- anonymous

The equation of a circle is x2 + y2 + Cx + Dy + E = 0. If the radius of the circle is decreased without changing the coordinates of the center point, how are the coefficients C, D, and E affected?
i dont get how to work with circles

- anonymous

Do you know how to complete the square?

- anonymous

no

- anonymous

Ok, completing the square is kind of a technique that is similar to factoring (somewhat). It's based on the formula \[(a\pm b)^2=a^2\pm2ab+b^2\] The idea is that you're trying to get a quadratic into a form that can become a perfect square. Before we work on your question, let's work with the idea first.\[x^2-4x+5=10\] Now, of course, the easiest thing would be to factor, but let's try completing the square instead.\[x^2-4x+\_\_=5\] You're trying to find the number that should go there to make it a perfect square trinomial.\[x^2-4x\color{red}{+4}=5\color{red}{+4}\]We added 4 on the left side to make it the perfect square trinomial (this is to make it fit the form above). However, to balance the equation, we need to add 4 to the right side as well.\[(x-2)^2=9\]That is called completing the square. Does that make sense so far?

- anonymous

kind of, yes

- anonymous

okay now what?

- anonymous

Alright, so what we need to do is complete the square for the equation of the circle above.
\[x^2+y^2+Cx+Dy+E=0\]\[(x^2+Cx+\_\_)+(y^2+Dy+\_\_)=-E\]\[\left(x^2+Cx+\left(\frac{C}{2}\right)^2\right)+\left(y^2+Dy+\left(\frac D2\right)^2\right)=-E+\left(\frac C2\right)^2+\left(\frac D2\right)^2\]\[\left(x+\frac C2\right)^2+\left(y+\frac D2\right)^2=-E+\left(\frac C2\right)^2+\left(\frac D2\right)^2\]This should look familiar from the form above.\[(x-h)^2+(y-k)^2=r^2\]where \(h=-\frac C2\), \(k=-\frac D2\), \(r^2=-E+\left(\frac C2\right)^2+\left(\frac D2\right)^2\)

- anonymous

If you're not quite 100% on completing the square yet, ^ is going to be a bit hard to follow.

- anonymous

okay i kinda follow its a bit hazy but its there

- anonymous

Alright, I'll go over completing the square in a sec again, but to finish up your question, the variables that affect the center of the circle are C and D. Your question asked that if r decreased, how would C, D, E be affected if the center of the circle didn't change. So, since the center didn't move, C and D stay constant. However, E needs to increase for the radius to decrease. Does that make sense? I'll go back to completing the square after this.

- anonymous

okay yeah that does

- anonymous

the only problem is thats not an answer

- anonymous

What is the answer according to your source?

- anonymous

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- anonymous

these are our options

- anonymous

With the way the question is written, I don't think any of the choices are correct. Are you sure that's the exact wording of the question?

- anonymous

yeah i copy and paste it

- anonymous

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- anonymous

I just double checked everything on my paper. I think their question is just straight up wrong. If it were - E or something instead of + E, then the choice would be on there. I'd just go with the last option since it's the closest one, but...wow...

- anonymous

i chose e

- anonymous

http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B6x%2B6y%2B5%3D0
http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B6x%2B6y%2B10%3D0
Also, ^ that is why I'm sure that their question is flawed.

- anonymous

The +10 one is clearly smaller than the +5 one, and the centers stayed the same.

- anonymous

Sorry for all the confusion.

- anonymous

its okay thanks tho

- anonymous

np

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