Confused about the equation of a cirlce

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GogetaORvegito?
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#1
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#1
The equation of a circle is always seen in this form:
(x - a)² + (y - b)² = r²

My question is, the subtraction means the circles position is moving to the right of the axis and up the y axis right. So in other words the circle is never travelling into the negatives?
Last edited by GogetaORvegito?; 1 year ago
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Always_Confused
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#2
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#2
Hello. I'm not sure what you mean by the circle never travels to the negatives.
'a' and 'b' in the equation are the x and y coordinates of the centre of a circle (a,b).
The centre of a circle could lie in a negative plane, i.e both a and b are negative.
My teacher had us derive the equation of a circle using Pythagoras' theorem, and I've always just thought of the (x-a) and (y-b) terms coming from that really.
I'm not sure whether that helps though!
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RogerOxon
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#3
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#3
(Original post by GogetaORvegito?)
The equation of a circle is always seen in this form:
(x - a)² + (y - b)² = r²

My question is, the subtraction means the circles position is moving to the right of the axis and up the y axis right. So in other words the circle is never travelling into the negatives?
Draw a diagram of a circle, centred on the origin. Note that x^2+y^2=r^2 is just Pythagoras' theorem, i.e. the distance from the origin to any point on the circle is r.

Now consider what the a and b offsets do - they're basically substitutions, e.g. X=x-a, Y=y-b, which moves the centre of the circle to (a,b). If a and b are both greater than, or equal to, r, then the circle will be entirely in the upper right quadrant, i.e. have no negative x or y values.
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RogerOxon
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#4
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#4
(Original post by GogetaORvegito?)
My question is, the subtraction means the circles position is moving to the right of the axis and up the y axis right. So in other words the circle is never travelling into the negatives?
a and b might be negative, or less than r.
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GogetaORvegito?
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#5
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#5
(Original post by Always_Confused)
Hello. I'm not sure what you mean by the circle never travels to the negatives.
'a' and 'b' in the equation are the x and y coordinates of the centre of a circle (a,b).
The centre of a circle could lie in a negative plane, i.e both a and b are negative.
My teacher had us derive the equation of a circle using Pythagoras' theorem, and I've always just thought of the (x-a) and (y-b) terms coming from that really.
I'm not sure whether that helps though!
Thanks for this
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GogetaORvegito?
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#6
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(Original post by RogerOxon)
a and b might be negative, or less than r.
That makes sense. I was just confusing myself before. Thanks
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GogetaORvegito?
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#7
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#7
(Original post by RogerOxon)
a and b might be negative, or less than r.
So if I were to draw the circle with equation ( x - 1 )² + ( y - 3 )² = 45 , will the circle have the radius of root 45 and go up 3 from the center and shift one value to the right
Last edited by GogetaORvegito?; 1 year ago
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RogerOxon
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#8
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(Original post by GogetaORvegito?)
So if I were to draw the circle with equation ( x - 1 )² + ( y - 3 )² = 45 , will the circle have the radius of root 45 and go up 3 from the center and shift one value to the right
Yes - its centre would be at (1, 3).
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GogetaORvegito?
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#9
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#9
(Original post by RogerOxon)
Yes - its centre would be at (1, 3).
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