# Math Help - graphing a circle

1. ## graphing a circle

Hello! I have the following equation to graph.

4x^2+y^2+8x-y+7=0.

I know that this is a circle but I only know how to graph it when there are simply a y^2 and x^2. The other variables are confusing me..what should I do? Thank you!

i also just figured out that the radius is 1 and the center (1,2) but what will the "4" do to the circle?

2. i also just figured out that the radius is 1 and the center (1,2) but what will the "4" do to the circle?

3. Originally Posted by ss103
Hello! I have the following equation to graph.

4x^2+y^2+8x-y+7=0.

I know that this is a circle but I only know how to graph it when there are simply a y^2 and x^2. The other variables are confusing me..what should I do? Thank you!

i also just figured out that the radius is 1 and the center (1,2) but what will the "4" do to the circle?
recheck your original equation ... it's not a circle.

The equation is in the form of an ellipse, but the +7 at the end is a problem.

4. but if you complete the square, you achieve this equation:

4(x-1)^2+(y-2)^2=1

which is of a circle...i just don't know what to do with the 4.

5. Originally Posted by ss103
but if you complete the square, you achieve this equation:

4(x-1)^2+(y-2)^2=1

which is of a circle...i just don't know what to do with the 4.
No, as you were told before, it is NOT a circle, it is an "ellipse". You can write that in the "standard" form for an ellipse: $\frac{(x-x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b}^2= 1$; an ellipse with center at $(x_0,y_0)$ and "semi-axes" along the x and y axes of lengths a and b. It is a circle if and only if the two "semi-axes" are the same: the radius of the circle.

$\frac{(x- 1)^2}{\left(\frac{1}{2}\right)^2}+ \frac{(y- 2)^2}{1^2}= 1.$

The center is at (1, 2) and the semi-axes are of lengths 1/2 and 1. Draw horizontal lines (parallel to the x-axis) from (1, 2) of length 1/2 ( from (1, 2) to (1/2, 2) and (3/2, 2)) and draw vertical lines (parallel to the y-axis) from (1, 2) of length 1 (from (1, 2) to (1, 1) and (1, 3)). If you like you can draw a rectangular box with center at (1, 2) and sides y= 1, y= 3, x= 1/2, x= 3/2. The ellipse will just fit into that box.

6. Originally Posted by ss103
but if you complete the square, you achieve this equation:

4(x-1)^2+(y-2)^2=1

which is of a circle...i just don't know what to do with the 4.
your original expression is incorrect. you originally posted ...

4x^2+y^2+8x-y+7=0.
or did you mean this ? ...

$4x^2+y^2+8x-\textcolor{red}{4}y+7=0$

and , one more time, it's not a circle ... it's an ellipse.

$\frac{(x-1)^2}{\left(\frac{1}{2}\right)^2} + \frac{(y-2)^2}{1^2} = 1$

semi-minor axis = $\frac{1}{2}$

semi-major axis = $1$

7. ohhh yesss i am sooo sorry about that !! but i don't understand where the 1/2 comes from in the denominator?

8. Originally Posted by ss103
ohhh yesss i am sooo sorry about that !! but i don't understand where the 1/2 comes from in the denominator?
it's the "4" in the numerator.

9. right so what does the 4 do to the graph of the circle...this is where i was initially confused :/

10. Originally Posted by ss103
right so what does the 4 do to the graph of the circle...this is where i was initially confused :/
it makes the conic an ellipse.