Confused about the equation of a cirlce

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?

(edited 3 years ago)

Reply 1

Always_Confused

13

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!

Reply 2

RogerOxon

21

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.

Reply 3

RogerOxon

21

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.

Reply 4

GogetaORvegito?

OP

13

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

Reply 5

GogetaORvegito?

OP

13

Original post by RogerOxon

a and b might be negative, or less than r.

That makes sense. I was just confusing myself before. Thanks

Reply 6

GogetaORvegito?

OP

13

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

(edited 3 years ago)

Reply 7

RogerOxon

21

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