## anonymous one year ago What is the standard form of the equation for this circle?

1. anonymous

2. anonymous

@hotguy @Vocaloid @ any one who can help me

3. 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?

4. anonymous

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

5. anonymous

right?

6. anonymous

Yup. Great job!

7. anonymous

thanks!

8. anonymous

np :)

9. anonymous

@Calcmathlete could you help me with another question?

10. anonymous

Sure

11. 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

12. anonymous

Do you know how to complete the square?

13. anonymous

no

14. 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?

15. anonymous

kind of, yes

16. anonymous

okay now what?

17. 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$$

18. anonymous

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

19. anonymous

okay i kinda follow its a bit hazy but its there

20. 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.

21. anonymous

okay yeah that does

22. anonymous

the only problem is thats not an answer

23. anonymous

What is the answer according to your source?

24. anonymous

25. anonymous

these are our options

26. 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?

27. anonymous

yeah i copy and paste it

28. anonymous

29. 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...

30. anonymous

i chose e

31. 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.

32. anonymous

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

33. anonymous

Sorry for all the confusion.

34. anonymous

its okay thanks tho

35. anonymous

np