an equation for a circle with center (0,0) and radius is 8
x^2+y^2=8^2
if centre is (a,b) and radius is r equation is given by... \[(x-a)^{2}+(y-b)^{2}=r^{2}\] try it out with the above values!!1
well the options are a)x^2+y^2=64 or b)x^2+y^2=8
you substitute the values in above equation and get answer.......
Actually if the center is (0,0), then the equation of a circle is x^2 + y^2 = r^2
soooo would the answer be x^2+y^2= 64 ?
Yup
thanks :)
okay so theres a circle, and the raduis = 7. and it says write an equation, would it be x^2+y^2=7^2 ?
They didn't give you the coordinates of its center point?
The equation of a circle is x^2 + y^2 = r^2 if and only if the the coordinates of its center point is (0,0).
Okay yes, thank you so much ! :)
But you didn't answer my question
oh, yeeh the coordinates are (0,0) the center
Okay, well, then, yes, your equation is correct. You can either say: x^2 + y^2 = 7^2 or x^2 + y^2 = 49
okay, and theres another circle, and the points are outside the circle, it says (6,1) how would I write that equation ?
You can use the formula the other guy provided
(x-a)^2+(y-b)^2=r^2 ?
Yes, that :)
Be careful with the negative
so it would be (0-6)^2+(0-1)^2=r^2 ?
I had a feeling you were going to do that. You don't replace x with zero by the way
and when I solve it would be 37=r^2 which would then be r=6.1 ?
Oh then what do I do ?
No, you replaced x and y with zeroes which is forbidden.
The center point is (a,b) not (x,y) in this case. In other words, you leave x and y alone
Wait a sec...The points are OUTSIDE the circle? You should probably post or draw the picture.
This is a different case if a point is outside the circle.
ohh okay hold on
Usually, the given formulas are used when a point is on the circle and you know the center point.
|dw:1341563486013:dw|
something like that
And (6,1) is not on the circle?
Can you figure out what any one of the points I just drew are? |dw:1341563615294:dw|
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