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new equation circle
- Thread starter czagara
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JuicyBurger
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\(\displaystyle (x-\frac{1}{2})^2 + (y-4)^2 = 21\)
This is already in the standard form of a circle, so I'm assuming all you need to do is graph it on your online program? (correct me if I'm wrong, but you didn't provide any instructions).
In this case this is the exact same thing we have been working on (in your other thread). Recall the equation of a circle
\(\displaystyle (x-h)^2 + (y-k)^2 = r^2\)
and what the constants h,k and r stand for?
Try this out.
This is already in the standard form of a circle, so I'm assuming all you need to do is graph it on your online program? (correct me if I'm wrong, but you didn't provide any instructions).
In this case this is the exact same thing we have been working on (in your other thread). Recall the equation of a circle
\(\displaystyle (x-h)^2 + (y-k)^2 = r^2\)
and what the constants h,k and r stand for?
Try this out.
czagara said:(x ? 1/2)2 + (y ? 4)2 = 21
Ok this one is even more confusing to me than the last one.
Where it the center? What is the radius?
This is ALL of the information you need to draw the graph.
czagara said:i thought this was already in general form???
Yep...it is.
And in the general form for the equation of a circle,
(x - h)[sup:3dlfu9tz]2[/sup:3dlfu9tz] + (y - k)[sup:3dlfu9tz]2[/sup:3dlfu9tz] = r[sup:3dlfu9tz]2[/sup:3dlfu9tz]
the center (h, k) and the radius (r) are right there.
JuicyBurger
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czagara said:this says it should be equal to 0
So in this case just subtract 21 from both sides to get the function set equal to zero:
\(\displaystyle (x-\frac{1}{2})^2+(y-4)^2 -21 = 0\)
JuicyBurger
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czagara said:apparently the computer saya that is not the answer
Can you give us a few more details on what it is asking?
Mrspi said:czagara said:i thought this was already in general form???
Yep...it is.
And in the general form for the equation of a circle,
(x - h)[sup:7evr59r2]2[/sup:7evr59r2] + (y - k)[sup:7evr59r2]2[/sup:7evr59r2] = r[sup:7evr59r2]2[/sup:7evr59r2]
the center (h, k) and the radius (r) are right there.
I'm sorry...I misunderstood. The equation you're given in is "standard form"
General form for the equation of a circle would be the result of expanding the squares, and getting the right side of the equation equal to 0.
Here's an example:
(x - 1)[sup:7evr59r2]2[/sup:7evr59r2] + (y + 3)[sup:7evr59r2]2[/sup:7evr59r2] = 25
Expand the squares:
x[sup:7evr59r2]2[/sup:7evr59r2] - 2x + 1 + y[sup:7evr59r2]2[/sup:7evr59r2] + 6x + 9 = 25
x[sup:7evr59r2]2[/sup:7evr59r2] - 2x + y[sup:7evr59r2]2[/sup:7evr59r2] + 6x + 10 = 25
Get the right side to be 0 by subtracting 25 from both sides:
x[sup:7evr59r2]2[/sup:7evr59r2] - 2x + y[sup:7evr59r2]2[/sup:7evr59r2] + 6y - 15 = 0
Now...you may need to "re-order" the terms to get your "online computer teacher" to accept this answer.
JuicyBurger
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Mrspi said:I'm sorry...I misunderstood. The equation you're given in is "standard form"
It seems we both misunderstood, and I agree, real teacher's are much more efficient and helpful than online teachers and tutors.
JuicyBurger
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czagara said:x^2-2x+y^2+4y-21=0 is not the answer
isn't that right?
Careful, what happens when you expand each of the squares? I got:
\(\displaystyle x^2 -x+\frac{1}{4}\) and...
\(\displaystyle y^2 - 8y +16\)
which gives us...
\(\displaystyle x^2 -x+\frac{1}{4}+y^2 - 8y +16 -21 = 0\)
I'm sorry, but I don't really know what you mean by "standard equation". It will be useful to me if you send the definition.czagara said:I'm sorry i know you guys are just as frustrated as i am. I will not be taking anymore math online classes ever again.
Do you have any more suggestions. I really appreciate everyone's help tonight.
(I know the equation
ax[sup:16zf8ufy]2[/sup:16zf8ufy]+by[sup:16zf8ufy]2[/sup:16zf8ufy]+cx+dy+f=0
is the "general" form).The equation
(x-h)[sup:16zf8ufy]2[/sup:16zf8ufy]+(y-k)[sup:16zf8ufy]2[/sup:16zf8ufy]=r[sup:16zf8ufy]2[/sup:16zf8ufy]
That is obtained completing the square, as you did, allows to realize that this is a circle,where h and k are the coordinates of the center (h,k) and r is the radius of the circle
I did not check the equation you wrote, but this may be the answer to write the standard equation, and the other equation
(x-h)[sup:16zf8ufy]2[/sup:16zf8ufy]+(y-k)[sup:16zf8ufy]2[/sup:16zf8ufy]=r[sup:16zf8ufy]2[/sup:16zf8ufy]
is what you need to graph the circle.