Question 253258
{{{4x^2+4y^2-24+16y-100=0}}}

{{{4x^2+4y^2-24+16y-100=0}}}

{{{4x^2+4y^2+16y-124=0}}}
<pre><font size = 4 color = "indigo"><b>
First of all notice that every term is divisible by 4,
so let's divide every term by 4:

{{{x^2+y^2+4y-31=0}}}

Let's add 31 to both sides:

{{{x^2+y^2+4y=31}}}

Let's complete the square of the last two terms {{{y^2+4x}}}
on the left side.  We take the coefficient of y, which is 4, 
multiply it by {{{1/2}}}, getting 2, then we square 2, and 
{{{2^2}}} is 4, so we add {{{""+4}}} to both sides:

{{{x^2+y^2+4y}}}{{{red("+4")=31}}}{{{red("+4")}}}

Let's put parentheses around the last three terms on the left
side and combine the terms on the right:

{{{x^2+(y^2+4y+4)=35}}}

Now factor the trinomials in the parentheses:

{{{x^2+(y+2)(y+2)=35}}}

That can be written as

{{{x^2+(y+2)^2=35}}}

In fact you could have skipped the step before that.

The standard form for a circle is

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

Since you didn't have to complete the square on the
x term, because you only had an {{{x^2}}} term and
no x-term, then to get it in the standard form you
have to write {{{x^2}}} as {{{(x-0)^2}}} to have the
standard form, so the answer is

{{{(x-0)^2+(y+2)^2=35}}}

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

is the equation of a circle

which has center (h,k) and radius r

{{{(x-0)^2+(y+2)^2=35}}}

is the equation for a circle which has center

(0,-2) and radius {{{sqrt(35)}}} which is about 5.92

and has this graph:

{{{drawing(200,200,-7,7,-9,5, graph(200,200,-7,7,-9,5),
line(0+.1,-2,0-.1,-2), line(0,-2+.1,0,-2-.1),line(0+.1,-2+.1,0-.1,-2-.1),line(0+.1,-2-.1,0-.1,-2+.1), locate(0,-2,"(0,-2)"),
circle(0,-2,sqrt(35)) )}}}

----------------------------------------------

Are you sure there wasn't supposed to be an x after the -24?
The reason I ask that is I don't think when you are advanced 
enough to be studying circles, that your teacher would 
still be testing you on such an elementary thing as to see 
whether you could combine the terms -24 and -100.  So I'm going 
to assume you really meant:

{{{4x^2+4y^2-24red(x)+16y-100=0}}}

First of all notice that every term is divisible by 4,
so let's divide every term by 4:

{{{x^2+y^2-6x+4y-25=0}}}

Let's group the the two terms in x first and the two terms
in y second, and add 25 to both sides:

{{{x^2-6x+y^2+4y=25}}}

To complete the square of the first two terms {{{x^2-6x}}}
we take the coefficient of x, which is -6, multiply it by {{{1/2}}},
getting -3, then we square -3, and {{{(-3)^2}}} is 9, so we add {{{""+9}}}
to both sides:

{{{x^2-6x}}}{{{red("+9")+y^2+4y=25}}}{{{red("+9")}}}

Let's put parentheses around the first three terms and
combinethe terms on the right.

{{{(x^2-6x+9)+y^2+4y=34}}}

Now lwt's complete the square of the last two terms {{{y^2+4x}}}
on the left side.  We take the coefficient of y, which is 4, 
multiply it by {{{1/2}}}, getting 2, then we square 2, and 
{{{2^2}}} is 4, so we add {{{""+4}}} to both sides:

{{{(x^2-6x+9)+y^2+4y}}}{{{red("+4")=34}}}{{{red("+4")}}}

Let's put parentheses around the last three terms on the left
side and combine the terms on the right:

{{{(x^2-6x+9)+(y^2+4y+4)=38}}}

Now factor the trinomials in the two parentheses:

{{{(x-3)(x-3)+(y+2)(y+2)=38}}}

That can be written as

{{{(x-3)^2+(y+2)^2=38}}}

In fact you could have skipped the step before that.

That's the standard form for a circle 

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

which has center (h,k) and radius r

{{{(x-3)^2+(y+2)^2=38}}}

is the equation for a circle which has center

(3,-2) and radius {{{sqrt(38)}}} which is about 6.16

and has this graph:

{{{drawing(200,200,-4,10,-9,5, graph(200,200,-4,10,-9,5),
line(3+.1,-2,3-.1,-2), line(3,-2+.1,3,-2-.1),line(3+.1,-2+.1,3-.1,-2-.1),line(3+.1,-2-.1,3-.1,-2+.1), locate(3,-2,"(3,-2)"),
circle(3,-2,sqrt(38)) )}}} 

Edwin</pre>