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cartesian equation circle question

can someone help me solve this??
:show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius
Original post by Carokelly123
can someone help me solve this??
:show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius


Is that the whole question
Original post by Carokelly123
can someone help me solve this??
:show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius


What circle?
Original post by 123Master321
Is that the whole question


yeah it is!
Original post by Carokelly123
yeah it is!


Not enough information to define a circle.
Original post by RDKGames
Not enough information to define a circle.


Let me math a sec
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B_9710
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Original post by Carokelly123
can someone help me solve this??
:show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius


Note that y/x=t y/x=t . Sub that expression for t into one of the expressions for x or y and then rearrange and you get the equation for a circle.
It's just converting from parametric to Cartesian.
Original post by Carokelly123
can someone help me solve this??
:show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius


(11+t2,t1+t2)=(X,Y)(\frac{1}{1+t^2}, \frac{t}{1+t^2}) = (X, Y)

Now you have two equations:

X=11+t2,X = \frac{1}{1+t^2},
Y=t1+t2Y = \frac{t}{1+t^2}

Rearrange one or both of the equations to get a function for t in terms of either X. Y or both, then substitute that function for t in the other. If the resulting equation is of the form:

(Y−a)2+(X−b)2=r2(Y-a)^2+(X-b)^2 = r^2

Then you have a circle, centre (a,b)(a,b) and radius rr

EDIT: Note there is an easy method for this particular example to do the simultaneous equations.
(edited 7 years ago)
t=y/x, substitute in x=1/(1+t^2) and rearrange to get x^2 - x +y^2 = 0, then complete the square to get (x-1/2)^2 + y^2 = (1/2)^2, i.e. a circle centered on (1/2, 0) with a radius of 1/2.
Original post by RogerOxon
t=y/x, substitute in x=1/(1+t^2) and rearrange to get x^2 - x +y^2 = 0, then complete the square to get (x-1/2)^2 + y^2 = (1/2)^2, i.e. a circle centered on (1/2, 0) with a radius of 1/2.


Please, no full solutions.

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