# Thread: Conics, finding the equation of a circle

1. ## Conics, finding the equation of a circle

I thought of posting my questions in already created thread with a similar topic but then I thought twice, thinking that it wouldn't be easier on the mods and it would just be invading other peoples threads.

I have many questions, many. I'm reading over the notes before starting the 40 questions that are to be done by the end of the day.

Okay so this is an example of switching the equation of a circle from standard to general.
It writes the following:
(x-root2)^2+(x-root2)^2=9
x^-root2x-root2x+2+y^2+2y+1=9
x^2-2root2x+y^2+2y+1=9

Okay, so I was trying this out if i could do it on my own and I think that they don't show all the steps like the foiling and what not, but where does the y come from?

Next question (it takes a while to get an answer so I should post more than 1 question, I feel.) Oh and by the way I'm sorry for not being able to do the root sign, the laptop does not have the #keypad so I can't use any alt key functions.

Is this a equation of a circle?
x^2+y^2-4y-4=0 yes..why?

2x^2+2y^2+4x-12y+10=0 yes, (remember to divide the equation by zero) why is a circle? How?

2x^2+y^2+7x-2y=0 no. How come?

2. Originally Posted by etheen
Is this a equation of a circle?
x^2+y^2-4y-4=0 yes..why?
The equation of a circle is in this form:
$(x - h)^{2} + (y - k)^{2} = r^{2}$
Where r is the radius and (h, k) is the center. You need to get your equation into this form by completing the square:
$x^{2} + y^{2} - 4y - 4 = 0$
$x^{2} + y^{2} - 4y = 4$
$x^{2} + y^{2} - 4y + 4 = 4 + 4$
$x^{2} + (y - 2)^{2} = 8$
So this is a circle with radius sqrt(8) and center at (0, 2).
2x^2+2y^2+4x-12y+10=0 yes, (remember to divide the equation by zero) why is a circle? How?
You probably mean divide by 2.
$2x^{2} + 2y^{2} + 4x - 12y + 10 = 0$
$x^{2} + y^{2} + 2x - 6y + 5 = 0$
Can you figure out the rest for this one?

01

3. ## parabola questions

No I mean, WHY is it a circle and why is t not? like what makes it not be a circle?
I don't know that.....
btw I did indeed mean divide by 2. Sorry.

Also, I hav parabola questions, I did all my math but i only have the parabola questions left.
Like, ind the focus of the parabola :
x^2-10x+25=(y-5)

but i have 1 conics question:
write the conic form of an ellipse determined by:
40x^2+5y-120=0 I got up to x^2/5 +y^2/40=120....but isn't it supposed to equal 1?

Also I must find the coordinates of the foci, the lenght of the major axis and the lenght of the minor axis. but i stoped in the begining because it wasnt equal to 1.

4. Originally Posted by etheen
No I mean, WHY is it a circle and why is t not? like what makes it not be a circle?
I don't know that.....
As I said, the equation of a circle has to be in this form: $(x - h)^{2} + (y - k)^{2} = r^{2}$. If you were to expand the binomials out you would notice that both the x^2 and y^2 terms have a leading coefficient of 1. With the equation 2x^2+2y^2+4x-12y+10=0, you can divide both sides by two (making the coefficients of x^2 and y^2 both be 1), and then you can complete the square. 2x^2+y^2+7x-2y=0 is not a circle because the coefficients of x^2 and y^2 are different and there's no way to make both of them 1.

Also, I hav parabola questions, I did all my math but i only have the parabola questions left.
Like, ind the focus of the parabola :
x^2-10x+25=(y-5)
Write the equation of the parabola in the form $(x-h)^2 = 4p(y-k)$, where (h, k) is the vertex and p is the directed distance from the vertex to the focus.
$x^2-10x+25=(y-5)$
$(x-5)^2=(y-5)$
Vertex is (5, 5), and 4p = 1, so p = 1/4. This parabola opens upwards, so the focus is (h, k+p), or (5, 5.25).

but i have 1 conics question:
write the conic form of an ellipse determined by:
40x^2+5y-120=0 I got up to x^2/5 +y^2/40=120....but isn't it supposed to equal 1?

Also I must find the coordinates of the foci, the lenght of the major axis and the lenght of the minor axis. but i stoped in the begining because it wasnt equal to 1.
Double-check this equation: 40x^2+5y-120=0. Shouldn't the y term be squared? Any terms missing?

01

5. Okay.
What is a circle?
Circle is locus of a point which moves such that its distance from a given fixed point is always a constant.
The fixed point is the centre of the circle and the constant distance is the radius.
Now let us attempt to find the equation of this locus.

Let the fixed point be $(h,k)$ and the constant distance be $r$.Let $(x,y)$ be any arbitrary point on the circle.

Distance between (h,k) and (x,y)= $\sqrt{(x-h)^2+(y-k)^2}$
Therefore, $\sqrt{(x-h)^2+(y-k)^2}=r$

On squaring we get $(x-h)^2+(y-k)^2=r^2.$

Every point on the circle will satisfy the above equation and therefore it can be said to be the equation of a circle having centre as $(h,k)$ and radius $r$.

Therefore, whenever you get an equation of the circle you are required to bring the equation in the above form and then will you be able to get the radius of the circle.

As to why a number is to be added or subtracted to the equation supplied to you.i suggest you take up a a book in elementary algebra and revise the factorization part.