Where does the line <x,y>=<4t,3t> intersect the circle $x^2+y^2=25$?

From <x,y>=<4t,3t>
=>> x=4t and y=3t
Then, substitute x and y to $x^2+y^2=25$.
==>> $16t^2+9t^2=25$
==>> $t^2=1$
==>> $t=+1,-1$

At t=1,-1 is the intersection point?
And is this what the question asked?

2. Originally Posted by noppawit
Where does the line <x,y>=<4t,3t> intersect the circle $x^2+y^2=25$?

From <x,y>=<4t,3t>
=>> x=4t and y=3t
Then, substitute x and y to $x^2+y^2=25$.
==>> $16t^2+9t^2=25$
==>> $t^2=1$
==>> $t=+1,-1$

At t=1,-1 is the intersection point?
And is this what the question asked?
Points given in Cartesian coordinates are probably required.

t = 1 => (4, 3).

t = -1 => (-4, -3).

The answer is easily understood if you draw a graph of the circle and the line.