need help with the general form of an circle ...

alexm544

alright, so i'm having some trouble with the general form of a circle. i know that it's basically
x²+y²+ax+by+c=0 , but this problem is giving me a hard time...

1. The problem statement, all variables and given/known data

a circle has the equasion 4x²+4y²+24x-16y-48=0. graph the cirlce using the center (h,k) and radius R. find the intercepts, if any.

... so, don't you combine all the like terms together first? so it would be:

(4x²+24x)²+(4y²-16y²)=48

but then what? or did i already screw it up?

Dick

Already screwed it up. You had 4x^2+24x and you just changed it into (4x^2+24x)^2. It's not like they are equal or anything. Start with 4x^2+24x=4(x^2+6x). You want to express x^2+6x in the form (x+a)^2-a^2 for some value of a. Expand (x+a)^2-a^2 and tell me what a should be. (This is called completing the square).

alexm544

wait, you totally lost me. where did 'a' come from?

Dick

I want to express that equation as a sum of squares. So I want to express x^2+6x in the form of a square. (x+a)^2-a^2 is EQUAL to x^2+6x for some choice of 'a'. I want you to figure out what that 'a' is and the show that they are equal. Multiply the square out and equate coefficients of the 'x' term.

statdad

When you have product like these

[tex]
\begin{align*}
(x+a)^2 & = x^2 + 2ax + a^2\\
(x-a)^2 & = x^2 - 2ax + a^2
\end{align*}
[/tex]

where [tex] a [/tex] may be a number, as in your problems, or another variable,
we refer to them as perfect squares . As examples

[tex]
\begin{align*}
(x-5)^2 & = x^2 - 2(5)(x) + 5^2 = x^2 - 10x + 25\\
(x+6)^2 & = x^2 + 2(6)(x) + 6^2 = x^2 + 12x + 36
\end{align*}
[/tex]

When we "complete the square" we are, in a sense, starting with a portion of the right-hand side and recreating the left.

Suppose you have

[tex]
x^2 + 12x
[/tex]

and know that it is the first 2/3 of the expanded form of [tex] (x+a)^2 [/tex]. Two questions come up:
* What is the value of [tex] a [/tex]?
* How can we recreate the [tex] (x+a)^2 [/tex]?

Here is how: in the two perfect square formulae, the constant term is found with these steps:
1) Divide the coefficient of [tex] x [/tex] by [tex] 2 [/tex]
2) Square the result of step 1

The thing to remember when you use this process in an equation is this: when you add something to the left of the equation, exactly the same amount must be added to the left. Two examples.

First, use completing the square to simplify this equation.

[tex] x^2 + 10x = 4 [/tex]

Here are the steps.
[tex]
\begin{align}
x^2 + 10x & = 4 \\
x^2 + 10x + 5^2 & = 4+5^2 \\
(x+5)^2 & = 29
\end{align}
[/tex]

How did I decide to add [tex] 5^2 [/tex] to each side? Since [tex] {10}/2 = 5 [/tex], the guidelines given above say that this is the appropriate choice.

One more:
[tex]
\begin{align*}
4y^2 + 104y & = 30\\
4\left(y^2 + 26y \right) & = 30\\
y^2 + 26y & = \frac{30}{4} = 7.5\\
y^2 + 26y + 13^2 & = 7.5 + 169\\
(y+13)^2 & = 176.5
\end{align*}
[/tex]
\end{align*}


Here's how this relates to circles:
1) Group the [tex] x [/tex] terms and the [tex] y [/tex] terms
2) If the coefficients of [tex] x^2 [/tex] and [tex] y^2 [/tex] from the terms, and divide both sides of the equation by the coefficient
3) Complete the square on the [tex] x [/tex] terms, and [tex] complete the square on the [tex] y [/tex] terms, just as shown above
4) When you simplify the left side and add the numbers on the right, your equation will be in standard form

Mentallic

You need to change the form for a circle into -

[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex] : where the centre of the circle is (h,k) and radius r

e.g.

[tex]x^{2}-6x+y^{2}+3y-5=0[/tex]

[tex][x^{2}-6x+(\frac{-6}{2})^{2}]-(\frac{-6}{2})^{2}+[y^{2}+3y+(\frac{3}{2})^{2}]-(\frac{3}{2})^{2}=5[/tex]

[tex](x-3)^{2}+(y+\frac{3}{2})^{2}=5+3^{2}+(\frac{3}{2})^{2}[/tex]

[tex](x-3)^{2}+(y+\frac{3}{2})^{2}=\frac{65}{4}[/tex]

Therefore, it is a circle with centre [tex](3,-\frac{3}{2})[/tex] and with radius [tex]\frac{\sqrt{65}}{2}[/tex]