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Confused about the equation of a cirlce
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GogetaORvegito?
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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?
(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
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!
'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
(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?
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?
is just Pythagoras' theorem, i.e. the distance from the origin to any point on the circle is 
.Now consider what the a and b offsets do - they're basically substitutions, e.g.
, 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
(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?
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?
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GogetaORvegito?
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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!
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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GogetaORvegito?
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#6
(Original post by RogerOxon)
a and b might be negative, or less than r.
a and b might be negative, or less than r.
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GogetaORvegito?
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#7
(Original post by RogerOxon)
a and b might be negative, or less than r.
a and b might be negative, or less than r.
Last edited by GogetaORvegito?; 1 year ago
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RogerOxon
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#8
(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
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
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GogetaORvegito?
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#9
(Original post by RogerOxon)
Yes - its centre would be at (1, 3).
Yes - its centre would be at (1, 3).
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