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Math Help - Equation of a Circle. HELP!

  1. #1
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    Equation of a Circle. HELP!

    How do you write the equation of the circle in standard form? and what is the center and radius of it?

    4x^2 + 4y^2 + 24x + 16y + 43 = 0

    A step-by-step view would really help to understand for future problems.
    Thanks!
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  2. #2
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    Quote Originally Posted by meiyukichan View Post
    How do you write the equation of the circle in standard form? and what is the center and radius of it?

    4x^2 + 4y^2 + 24x + 16y + 43 = 0

    A step-by-step view would really help to understand for future problems.
    Thanks!
    Re-arange:
    4x^2 + 24x + 4y^2 + 16y + 43 = 0

    \Rightarrow 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0.

    Now complete the square in each bracket. You should see where to go from there.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Re-arange:
    4x^2 + 24x + 4y^2 + 16y + 43 = 0

    \Rightarrow 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0.

    Now complete the square in each bracket. You should see where to go from there.
    What do you mean by complete the square in each bracket? I don't understand...
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  4. #4
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    Quote Originally Posted by meiyukichan View Post
    What do you mean by complete the square in each bracket? I don't understand...
    I will continue from here 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0 (*)

    x^2+6x=x^2+6x+9-9=(x+3)^2 -9 (1)


    y^2+4y=y^2+4y+4-4=(y+2)^2-4 (2)

    Substitute (1) and (2) into (*):

    you get 4*(x+3)^2-4*9+4*(y+2)^2-4*4+43=4(x+3)^2+4(y+2)^2=9 Center is at x=-3 and y=-2 ,Radius is 3
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Re-arange:
    4x^2 + 24x + 4y^2 + 16y + 43 = 0

    \Rightarrow 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0.

    Now complete the square in each bracket. You should see where to go from there.
    Quote Originally Posted by meiyukichan View Post
    What do you mean by complete the square in each bracket? I don't understand...
    The first bracket is x^2 + 6x = (x + 3)^2 - 9. Get it?
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    The first bracket is x^2 + 6x = (x + 3)^2 - 9. Get it?
    Then is this correct?

    4 (x - 3)^2 - 9 + 4 (y + 2)^2 - 4 + 43 = 0

    Rearrange: 4 (x - 3)^2 + 4 (y + 2)^2 - 4 + 43 -9 = 0

    4 (x - 3)^2 + 4 (y + 2)^2 = -29


    Does that mean that the center would be (3,2)?
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  7. #7
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     <br />
4x^2 + 4y^2 + 24x + 16y + 43 = 0<br />

    Rearrange the equation:

    4x^2 + 24x + 4y^2 + 16y = -43

    Try and make two perfect squares on the left side of the eqn.

    4x^2 + 24x + 36 + 4y^2 + 16y + 16 = -43 + 36 + 16

    (2x+6)^2 + (2y+4)^2 = 9

    Then find the centre and radius of the circle
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  8. #8
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    Quote Originally Posted by meiyukichan View Post
    Then is this correct?

    4 {\color{red}[} (x {\color{red}+} 3)^2 - 9{\color{red}]} + 4 {\color{red}[} (y + 2)^2 - 4 {\color{red}]}+ 43 = 0

    Rearrange: 4 (x {\color{red}+}  3)^2 + 4 (y + 2)^2 - {\color{red}16} + 43 - {\color{red}36} = 0

    4 (x {\color{red}+}  3)^2 + 4 (y + 2)^2 = -29 Mr F says: This should have told you that there were mistakes ..... How can the square of the radius be less than zero?

    Does that mean that the center would be (3,2)?
    Note the corrections I have made in red. The correction of the other mistake in the last line is left for you.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Note the corrections I have made in red. The correction of the other mistake in the last line is left for you.
    Then I should get:
    4 (x + 3^2) + 4(y + 2)^2 = 3^2
    ...?

    With the center being (3,2) and the radius 3...?


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  10. #10
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    Quote Originally Posted by meiyukichan View Post
    Then I should get:
    4 (x + 3)^{\color{red}2} + 4(y + 2)^2 = 3^2
    ...?

    With the center being (3,2) and the radius 3...?

    Apart from the typo, you're almost correct. The centre is at (-3, -2).

    By the way, there was an insulting post which I have deleted. My apologies on behalf of MHF if you had already read it.
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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    Apart from the typo, you're almost correct. The centre is at (-3, -2).

    By the way, there was an insulting post which I have deleted. My apologies on behalf of MHF if you had already read it.
    Oh! Thank you. I forgot that the equation was like this (x - h)^2 + (y - k)^2 = r^2
    meaning that since it was a positive, it is a negative then...

    I didn't read the insulting post, but if it was something like how can I make such a simply mistake and saying how stupid I am, then I don't care because I'm too young to understand this stuff fully anyways...
    I'm only in the 8th grade...
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