Results 1 to 5 of 5

Math Help - Circle equation, centre and radius

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    1

    Circle equation, centre and radius

    Hi! I am new here and need some help. I am doing a maths course at home through Open Uni and I have got stuck.

    The equation

    x^2 + 10x + y^2 - 6y + 9 = 0

    represents a circle. Find the centre and it's radius.

    I have had a look around online at other examples and it says to put it in a certain format with h,k & r. But I just can't follow it to make the equation given here fit in. Can anyone give me some guidance and help working it out please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    Let C be the circle center A(x_A,y_A) and radius R

    A point M(x,y) is on the circle if and only if
    AM = R which is equivalent to AM^2 = R^2

    Going through coordinates
    (x-x_A)^2 + (y-y_A)^2 = R^2

    Your equation
    x^2 + 10x + y^2 - 6y + 9 = 0
    can be written as
    (x+5)^2 - 25 + (y-3)^2 - 9 + 9 = 0
    (x+5)^2 + (y-3)^2 = 25 = 5^2

    Now I think that you can find the center and the radius
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by Bunnykins87 View Post
    Hi! I am new here and need some help. I am doing a maths course at home through Open Uni and I have got stuck.

    The equation

    x^2 + 10x + y^2 - 6y + 9 = 0

    represents a circle. Find the centre and it's radius.

    I have had a look around online at other examples and it says to put it in a certain format with h,k & r. But I just can't follow it to make the equation given here fit in. Can anyone give me some guidance and help working it out please?
    A circle with the center C(h, k) and the radius r has the equation

    (x-h)^2+(y-k)^2=r^2

    I'll show you the steps to transform your equation into the form of the general equation of a circle:
    <br />
\begin{aligned}x^2 + 10x + y^2 - 6y + 9 &= 0 \\<br />
x^2 + 10x + y^2 - 6y &=- 9 \\<br />
x^2+10x+25 + y^2 - 6x +9 &=-9+25+9\\<br />
(x+5)^2+(y-3)^2&=5^2\end{aligned}

    Thus the given equation describes a circle with C(-5, 3) and r = 5
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by Bunnykins87 View Post
    I have had a look around online at other examples and it says to put it in a certain format with h,k & r. But I just can't follow it to make the equation given here fit in. Can anyone give me some guidance and help working it out please?
    If you are having trouble rewriting the equation in the appropriate form, you should familiarize yourself with the method of completing the square. That will make these types of problems easy, and it is very useful to know in general.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    1
    Quote Originally Posted by Bunnykins87 View Post
    Hi! I am new here and need some help. I am doing a maths course at home through Open Uni and I have got stuck.

    The equation

    x^2 + 10x + y^2 - 6y + 9 = 0

    represents a circle. Find the centre and it's radius.

    I have had a look around online at other examples and it says to put it in a certain format with h,k & r. But I just can't follow it to make the equation given here fit in. Can anyone give me some guidance and help working it out please?
    Hi, I'm in year 11 doing AS level maths.
    Complete the square on it so it becomes
    (x+5)^2 + (y-3)^2
    expand it to show the numbers at the end
    -> (5x5) +25 and (-3x-3) +3
    x^2 + 10x + 25 + y^2 - 6y + 9 = 25
    [the number after the = is 19 because the right side numbers add up to
    34 and 34-9 = 25]
    so r^2 = 25
    therefore, r = 5
    so therefore, the centre (h,k) = (-5, 3) [because x+5 = 0 so x must be 5 and y-3 = 0 so y must be 3)
    so h = -5 and k = 3

    [I think this is right]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Determine the centre and radius of a circle
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 10th 2011, 04:57 AM
  2. Replies: 1
    Last Post: September 15th 2010, 02:04 PM
  3. Replies: 6
    Last Post: July 8th 2010, 05:39 PM
  4. Replies: 2
    Last Post: April 8th 2010, 07:43 PM
  5. Replies: 7
    Last Post: May 24th 2009, 07:41 AM

Search Tags


/mathhelpforum @mathhelpforum