Conics - How to identify figures in the plane?

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
x(x - 3) + y² = 0
c = (3,0)
(x-3)² + y² = 0
(3-3)² + y² -3 = 0
y = 0
x² + 0² - 3x = 0
x(x-3) = 0
x = 3
y = 0
It's a circle, the two square letters are round perfectly.

I would like you to know my errors.

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tnich
Homework Helper

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
x(x - 3) + y² = 0
c = (3,0)
(x-3)² + y² = 0
(3-3)² + y² -3 = 0
y = 0
x² + 0² - 3x = 0
x(x-3) = 0
x = 3
y = 0
It's a circle, the two square letters are round perfectly.

I would like you to know my errors.
You are correct that it is a circle. You have not correctly found the center. For a circle, you should also find the radius. To do this, put the equation of the circle in this form:
$(x-a)^2 +(y-b)^2 = r^2$
This expresses the relationship that each point $(x,y)$ on the circle is a distance $r$ from the center $(a,b)$.

Math_QED
Homework Helper
2019 Award
From what I understand, you calculate points where the curve intersects with the x-axis.

This is of course not sufficient. If you want to show it is a circle, then show you can write it in the form

$(x-x_0)^2 + (y-y_0)^2 = r^2$

Hint: complete the square with respect to $y$

vela
Staff Emeritus
Homework Helper
x(x - 3) + y² = 0
(x-3)² + y² = 0
Your first mistake is this step: $(x-3)^2 \ne x(x-3)$.

(x - 3)² + (y - 0)² = 0

Your first mistake is this step: $(x-3)^2 \ne x(x-3)$.
Thanks a lot, Vela!

Math_QED
Homework Helper
2019 Award
(x - 3)² + (y - 0)² = 0
That can't be correct. A circle has a strictly positive radius.

Moreover, $(x-3)^2 + (y-0)^2 = x^2 + 9 - 6x + y^2$ is not what you were given.

tnich
Homework Helper
(x - 3)² + (y - 0)² = 0
If you multiply out $(x-3)(x-3)$, do you get $x^2-3x$?

(x - 3)² + (y - 0)² = 0
Did I hit now?

If you multiply out $(x-3)(x-3)$, do you get $x^2-3x$?
Ahhh... Thanks, Nich!

That can't be correct. A circle has a strictly positive radius.

Moreover, $(x-3)^2 + (y-0)^2 = x^2 + 9 - 6x + y^2$ is not what you were given.
My professor said that the equation doesn't need to do like that in the second member. So he is wrong? [thinking]

Folks,

I'm going to the college right now. I'll give you the solution in the night or even in the college if I access the computers from there.

Math_QED
Homework Helper
2019 Award
My professor said that the equation doesn't need to do like that in the second member. So he is wrong? [thinking]
I don't know exactly what you mean, but I just meant that if you expand whar you wrote down, you don't get what you started with, so it couldn't be correct.

tnich
Homework Helper
My professor said that the equation doesn't need to do like that in the second member. So he is wrong? [thinking]
Here's a plan. Take $(x-a)^2 +(y-b)^2 = r^2$
and multiply out $(x-a)^2$ and $(x-b)^2$. To get a polynomial in x and y: $x^2 + sx + t + y^2 + uy + v = r^2$. Then set the coefficient of $x$ ($u =$ ?) in this equation with coefficient of $x$ in your circle equation and solve for $a$. Then figure out what $r$ needs to be to make the constant term ($t + v - r^2$) equal to zero as it is in your circle equation.
Edit: In line with @Math_QED's comment, also make sure that all of the other coefficients match between the two equations.

Last edited:
Math_QED
Homework Helper
2019 Award
Here's a plan. Take $(x-a)^2 +(y-b)^2 = r^2$
and multiply out $(x-a)^2$ and $(x-b)^2$. To get a polynomial in x and y: $x^2 + sx + t + y^2 + uy + v = r^2$. Then set the coefficient of $x$ ($u =$ ?) in this equation with coefficient of $x$ in your circle equation and solve for $a$. Then figure out what $r$ needs to be to make the constant term ($t + v - r^2$) equal to zero as it is in your circle equation.
I wouldn't recommend this approach, as initially it isn't given that the curve is a circle.

tnich
Homework Helper
I wouldn't recommend this approach, as initially it isn't given that the curve is a circle.
True. On the other hand, if it gives a value for the center and gives a positive value for $r^2$, then it proves that the curve is a circle.

Mark44
Mentor
That can't be correct. A circle has a strictly positive radius.
Or a zero radius, for a so-called degenerate circle that consists of a single point. We can also have degenerate ellipses and degenerate hyperbolas, the latter of which consist of two lines.

LCKurtz
Homework Helper
Gold Member

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
In post #3 Math_QED gave the hint to complete the square with respect to $y$. I think he meant with respect to $x$. That should be your first step, and not doing that is what everyone is hinting at is causing your mistakes.

Circle: ~x² + ~y² = R²
(x² - 3)² x= x² - 6x +9
~y = y; x: x - 3, c = (3,0)
(x - 3)² + y² → x = 3
I think now it's right!

symbolipoint
Homework Helper
Gold Member

Homework Statement

Identify the following figures in the plane, saying if they are circles, ellipses, hyperboles or parabolas and giving their centers, rays, foci, and guidelines as the case may be:

x² + y² - 3x = 0

The Attempt at a Solution

x² + y² - 3x = 0
x(x - 3) + y² = 0
c = (3,0)
(x-3)² + y² = 0
(3-3)² + y² -3 = 0
y = 0
x² + 0² - 3x = 0
x(x-3) = 0
x = 3
y = 0
It's a circle, the two square letters are round perfectly.

I would like you to know my errors.
CIRCLE; because both x and y are to exponent 2, possibly including one or both of the variables to exponent 1. AND coefficient on the exponent-two variables is 1 each. Nothing more is needed for identifying the conic section shape.

CIRCLE; because both x and y are to exponent 2, possibly including one or both of the variables to exponent 1. AND coefficient on the exponent-two variables is 1 each. Nothing more is needed for identifying the conic section shape.
Symboli,

I restated, am I right now? It seems that gave the same result. Hm...

symbolipoint
Homework Helper
Gold Member
(x - 3)² + (y - 0)² = 0
CIRCLE; because both x and y are to exponent 2, possibly including one or both of the variables to exponent 1. AND coefficient on the exponent-two variables is 1 each. Nothing more is needed for identifying the conic section shape.
Okay, I was wrong. It might be a degenerate circle.

x^2-3x+y^2=0
-
x^2-3x+(3/2)^2+y^2=(3/2)^2
(x-3/2)^2+y^2=(3/2)^2

So I was correct the FIRST time. The shape IS a circle.
Center is ( 3/2, 0 ) and the radius is 3/2.

Okay, I was wrong. It might be a degenerate circle.

x^2-3x+y^2=0
-
x^2-3x+(3/2)^2+y^2=(3/2)^2
(x-3/2)^2+y^2=(3/2)^2

So I was correct the FIRST time. The shape IS a circle.
Center is ( 3/2, 0 ) and the radius is 3/2.
@symbolipoint,

Why 3/2 and not 3? I didn't understand.

symbolipoint
Homework Helper