equation of a circle

So.... in the context of a circle on a graph this would mean that the circle's center is at (1,-1) and it's radius is 9.

Remember the equation has \(\displaystyle x-h\) and \(\displaystyle y-k\), where h and k are the x and y-coordinates, respectively. This means that if you have \(\displaystyle (x-1)^2\) that the x-coordinate will be at \(\displaystyle 1\) not \(\displaystyle -1\).
 
JuicyBurger said:
So.... in the context of a circle on a graph this would mean that the circle's center is at (1,-1) and it's radius is 9.

Remember the equation has \(\displaystyle x-h\) and \(\displaystyle y-k\), where h and k are the x and y-coordinates, respectively. This means that if you have \(\displaystyle (x-1)^2\) that the x-coordinate will be at \(\displaystyle 1\) not \(\displaystyle -1\).
Click to expand...

You can graph this circle on graph paper easily. You know the center is at (1, -1). You know the radius is 9. Use a set of compasses...set the distance between the pencil point and the sharp point at 9 units. Put the sharp point on the center of the circle (1, -1) and draw the circle. That's it. That's the graph of the circle.
 
so plotting on a graph, i would plot (1, 0) for the x coordinate, and (0,1) for the y coordinate and do i also have to plug in the radius, and if so how to do you plot the radius as 9?
 
well i plotted 1, -1 as the center and went out 9 and that didn't work
 
czagara said:
so plotting on a graph, i would plot (1, 0) for the x coordinate, and (0,1) for the y coordinate and do i also have to plug in the radius, and if so how to do you plot the radius as 9?
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Well not exactly. The points on the circle are those points 9 units from the center...which is why I suggested using a compass...if you know the center, and you know the radius, set the distance between the pencil point and the sharp point at 9 units. Put the sharp point on the center of the circle. Draw the circle.
 
well i don't have any graph paper and i have to submit this answer over the internet so i have to use the supplied computer graph given to me.
 
czagara said:
well i plotted 1, -1 as the center and went out 9 and that didn't work
Click to expand...

I'm not quite sure how you will need to do this on your online program, so I can't help you there unless you give more details.
 
JuicyBurger said:
czagara said:
well i plotted 1, -1 as the center and went out 9 and that didn't work
Click to expand...

I'm not quite sure how you will need to do this on your online program, so I can't help you there unless you give more details.
Click to expand...

Ok...I don't know how you're expected to do this graph on your computer program, either.

BUT....
You know the center. You could find that. And then from the center, go UP 9 units, to find a point on the circle. From the center, go TO THE RIGHT 9 units to find another point on the circle. From the center go DOWN 9 units to find another point on the circle. And from the center, go TO THE LEFT 9 units to find still another point on the circle.

A smooth curve through the 4 points you've plotted should be the graph of the circle.

I am surely VERY old-fashioned. This is such a simple graph to do on graph paper.
 
Mrspi said:
I am surely VERY old-fashioned. This is such a simple graph to do on graph paper.
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Too true, I haven't seen a compass since I was quite young (back when paper was still used!) :D
 
czagara said:
Write the equation of the circle in standard form. Then sketch the circle.

x^2 + y^2 ? 2x + 2y ? 79 = 0

(x^2-2x+1) I am not really sure if i am doing this right and i don't really know what do to next.
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----------------------------

(x-1)^2 - 1 + (y+1)^2 -1 -79 = 0

(x-1)^2 + (y+1)^2 = 81

Center of circle (a,b)
a=-(-1) = 1
==>> C(1,-1)
b=-(+1) =-1

Raduis of circle is square root 81 which is 9.
 
czagara said:
I really am not understanding what h, k and r are?
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Do you know what the equation
(x-h)[sup:27o3bv2n]2[/sup:27o3bv2n]+(y-k)[sup:27o3bv2n]2[/sup:27o3bv2n]=r[sup:27o3bv2n]2[/sup:27o3bv2n]
means?
Remember that h and k are the coordinates of the center (h,k) and r is the radius of the center
If not, may be you should go back to your notes
 
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