Is this circle equation in general form solveable?

Hi,

I have the equation -16x^{2} -128x+12y^{2} -48y-400=0 that I need to get into standard form. It looks like it's a circle equation, but when I try to remove the coefficients in front of x and y, I find I'm unable to get them both "alone", or without coefficients. Without that, I can't get perfect squares and so I can't solve the problem. Is there some way to get rid of those coefficients so I can complete the square and solve the problem?

Thanks!

Re: Is this circle equation in general form solveable?

Quote:

Originally Posted by

**KevinShaughnessy** I have the equation $\displaystyle -16x^2 -128x+12y^2 -48y-400=0$ that I need to get into standard form. It looks like it's a circle equation.

First, that is not a circle, it is a hyperbola. Divide through by $\displaystyle (-16)(12).$

Re: Is this circle equation in general form solveable?

The first thing I would do it factor out the coefficients of $\displaystyle x^2$ and $\displaystyle y^2$ separately, then complete the squares:

$\displaystyle -16(x^2+ 8x)+ 12(y^2- 4y)= 400$

You can complete the square in the first parentheses by adding $\displaystyle (8/2)^2= 16$ and in the second $\displaystyle (4/2)^2= 4$.

Of course, the first would be multiplied by -16 and the second by 12 to add them on the right.

Re: Is this circle equation in general form solveable?

Re: Is this circle equation in general form solveable?

i totally agree with Plato. but do ypou know how and why.

any second degree equation it two variables can represent a circle, an ellipse or a hyperbola. How do find it be observation.

it the equation has no term containing xy. and

coefficient of x^2 is equal to coefficient to that of y^2 then its graph will be a circle.

If the coefficients of x^2 and y^2 are not equal but have the same sign then it will represent an Ellipse and otherwise a hyperbola.

Quote:

Originally Posted by

**Plato** First, that is not a circle, it is a hyperbola. Divide through by $\displaystyle (-16)(12).$

Re: Is this circle equation in general form solveable?

Quote:

Originally Posted by

**ibdutt** i totally agree with Plato. but do ypou know how and why.

any second degree equation it two variables can represent a circle, an ellipse or a hyperbola. How do find it be observation.

it the equation has no term containing xy. and

coefficient of x^2 is equal to coefficient to that of y^2 then its graph will be a circle.

If the coefficients of x^2 and y^2 are not equal but have the same sign then it will represent an Ellipse and otherwise a hyperbola.

Ibdutt

I am being reading your posts for more that 4 months and I got the impression that you are a good Indian Mathematician..

I believe that the comments you posted yesterday are not appropriate..

First of all you forgot that there is another curve called Parabola.....therefore the so called second degrees curves of otherwise called CONIC SECTIONS ( Study by the Greek Geometer Apollonius of Perga 262 BC - 190 BC ) are 4 and not 3 .

Second the critiria you have used to classify the different conics represented by the whole second degree equation of two variables x,y ( ax^2+2bxy+cy^2+2dx+2ey+f=0) are not correct..

I therefore advise you to revise a good manual of Analytic Geometry ....

Do you know what kind of curve the second degree equation : ax^2+2axycosθ+ay^2+2dx+2ey+f=0 does represent?

well you maybe not ...a CIRCLE.....BUT THE coordinate system you are refered to is not orthogonal.....

Thus if we consider the whole second degree equation for two variables x,y ax^2+2bxy+cy^2+2dx+2ey+f=0

it represents

1. A real ellipse if ac-b^2>0 and aD<0 where D is the determinant of the coefficients D= [a,b,d b,c,e d,e,f ]

2. A real parabola if ac-b^2= 0 and D not equal to zero

3. A real hyperbola if ac-b^2<0 and aD not equal to zero.

For a circle we need a = c and b = 0 only if we are reffering to an orthogonal coordinate system .

MINOAS

G. M & R.

Re: Is this circle equation in general form solveable?

MINOANMAN

Thanks for pointing out. In fact I did not wish to include the complete discussion on the topic. Any how thanks a million