Find the radius & co-ordinates of the centre of the circle

**Maths_noob** · May 23rd 2009, 05:06 AM

Hello, This is my first post so please be nice

Find the radius & co-ordinates of the centre of the circle for the following

x^2 + y^2 - 4x - 6y = 0

x^2 + y^2 - 11x + 8y -12 = 0

3x^2 + 3y^2 - 12x - 15y + 30 = 0

Please could someone link me to a guide of how to work them out or even go through the process of how to do it for me

Any Help is very must appreciated
Thanks

**Opalg** · May 23rd 2009, 06:51 AM

Originally Posted by Maths_noob

Hello, This is my first post so please be nice

Find the radius & co-ordinates of the centre of the circle for the following

x^2 + y^2 - 4x - 6y = 0

x^2 + y^2 - 11x + 8y -12 = 0

3x^2 + 3y^2 - 12x - 15y + 30 = 0

Please could someone link me to a guide of how to work them out or even go through the process of how to do it for me

I'll do the first one, then you can do the others.

$x^2 + y^2 - 4x - 6y = 0$

Start by collecting the x terms together, then the y terms:

$x^2 - 4x + y^2 - 6y = 0$

Now complete the square in the x terms and the y terms:

$x^2 - 4x {\color{red}{}+4} + y^2 - 6y {\color{blue}{}+9} = {\color{red}4} {\color{blue}{}+9}$ ,
$(x-2)^2 + (y-3)^2 = 13$ .

Now you have to remember that the equation of a circle with centre (a,b) and radius r is $(x-a)^2 + (y-b)^2 = r^2$ . Compare that with the previous equation, and you see that this circle has centre (2,3) and radius $\sqrt{13}$ .

Do the other two problems in the same way. For the next one, you may want to start by shifting the constant term across to the other side of the equation. For the last one, you'll want to do the same thing, plus divide through by 3 before completing the square.

**Maths_noob** · May 24th 2009, 02:44 AM

$x^2 + y^2 - 11x + 8y - 12 = 0$

x terms together, then the y terms:

$x^2 - 11x + y^2 + 8y -12 = 0$

Complete the square

$x^2 - 11x + 11 + y^2 + 8y - 8 - 12 = 11 - 8$

$(x - a)^2 + (t - b)^2 = r^2$

$(x - 3.31662479)^2 + (y - 2.828427125)^2 = 3$

Would that be right? I think ive majorly cocked up somewhere tho lol

**yeongil** · May 24th 2009, 03:48 AM

Whoops, that's not right. Let's try again:
$x^2 + y^2 - 11x + 8y - 12 = 0$
Group like terms and move the 12 over:
$x^2 - 11x + y^2 + 8y = 12$
To complete the square, take the x coefficient, divide by 2, and square the result. Do the same for the y term:
$x^2 - 11x + \frac{121}{4} + y^2 + 8y + 16 = 12 + \frac{121}{4} + 16$
Factor and simplify:
$\left(x - \frac{11}{2}\right)^2 + (y + 4)^2 = \frac{233}{4}$

So the radius is $\sqrt{\frac{233}{4}}$ or $\frac{\sqrt{233}}{2}$ and the center is at $\left(\frac{11}{2}, -4\right)$ .

01

**Opalg** · May 24th 2009, 03:52 AM

Originally Posted by Maths_noob

$x^2 + y^2 - 11x + 8y - 12 = 0$

x terms together, then the y terms:

$x^2 - 11x + y^2 + 8y -12 = 0$

Complete the square

$x^2 - 11x + 11 + y^2 + 8y - 8 - 12 = 11 - 8$

$(x - a)^2 + (t - b)^2 = r^2$

$(x - 3.31662479)^2 + (y - 2.828427125)^2 = 3$

Would that be right? I think ive majorly cocked up somewhere tho lol Er, yes, you could say that.

When you complete the square, you divide the coefficient of x (or y, or whatever the variable is) and then square it, to get the constant term, which you then add to both sides of the equation.

$x^2 - 11x + y^2 + 8y -12 = 0$

$x^2 - 11x + y^2 + 8y =12$

$x^2 - 11x + {\color{red}\bigl(\tfrac{11}2\bigr)^2} + y^2 + 8y + {\color{blue}4^2} = 12 {\color{red}{} +\frac{121}4} {\color{blue}{}+ 16}$

Now in the next line the number that goes into the bracket is half the coefficient of x (or y, or whatever):

$(x - {\color{red}\tfrac{11}2})^2 + (y+{\color{blue}4})^2 = \frac{233}4$

So the centre is $\bigl(\tfrac{11}2,-4\bigr)$ and the radius is $\frac{\sqrt{233}}2$ .

Edit. See? yeongil and I are telling you the exact same thing, so it must be right.

**Maths_noob** · May 24th 2009, 04:22 AM

$3x^2 + 3y^2 - 12x - 15y + 30 = 0$

group terms:

$3x^2 -12x + 3y^2 - 15y = 30$

complete the square (the part i get confused lol)

$3x^2 - 12x + (12/3)^3 + 3y^2 - 15y + 7.5^2 = 11+ 11/2 + 56.25$

$(x - 12/3 )^2 + (y + 7.5)^2 = 289/4$

lol bet I've gone wrong but I had a go

**yeongil** · May 24th 2009, 04:47 AM

$3x^2 + 3y^2 - 12x - 15y + 30 = 0$
$3x^2 - 12x + 3y^2 - 15y = -30$ (not 30)

This one is different, because you have coefficients for x squared and y squared. First, factor out the 3:
$3(x^2 - 4x) + 3(y^2 - 5y) = -30$

Divide by 3:
$(x^2 - 4x) + (y^2 - 5y) = -10$

Complete the square:
$(x^2 - 4x + 4) + \left(y^2 - 5y + \frac{25}{4}\right) = -10 + 4 + \frac{25}{4}$

$(x - 2)^{2} + \left(y - \frac{5}{2}\right)^{2} = \frac{1}{4}$

So the radius is at $\frac{1}{2}$ and the center is at $\left(2, \frac{5}{2}\right)$ .

I also think you're going to have to review/re-learn completing the square.

Edit: And, I also think I have to watch my typos.

01

**Maths_noob** · May 24th 2009, 07:41 AM

Originally Posted by yeongil

I also think you're going to have to review/re-learn completing the square.

Yeah I reckon as well m8, is there any links you recommended to threads which will help?

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