Is there an equation for a straight line with a gradiant m through a unit circle (x^2 + y^2 = 1).
For example if you wanted to draw a straight line through this unit circle with an angle of pi/4 radians. pi/2 radians, etc (clearly the straight line for pi/2 radians would be x=0 but is there some formula for constructing such a line that works in every case?).
+23251 The line y = tan(pi/4)·x intersects the circle x2 + y2 = 1
at the points ( -sqrt(2)/2, -sqrt(2)/2 ) and ( sqrt(2)/2, sqrt(2)/2 ).
In general: the x-values of the points of intersection of the circle x2 + y2 = 1 and y = mx can be found by
replacing the value of y in the equation of the circle with the value mx ---> x2 + (mx)2 = 1
---> x2 + m2x2 = 1 ---> x2(1 + m2) = 1 ---> x2 = 1 / (1 + m2)
---> lower value: x = - sqrt( 1 / (1 + m2) )
---> upper value: x = sqrt( 1 / (1 + m2) )
+23251 The slope of a line is equal to the tangent of the angle that is formed by the line with the positive x-axis.
Thus, the slope of a line at an angle of pi/4 radians = tan(pi/4) = 1.
One other question...
What would the range of this function be, if you wanted to limit the length of the segment to 'fit' inside the circle.
\(x^2+y^2=1\)
\(y=\tan \left(\frac{\pi }{4}\right)x\)
+23251 The line y = tan(pi/4)·x intersects the circle x2 + y2 = 1
at the points ( -sqrt(2)/2, -sqrt(2)/2 ) and ( sqrt(2)/2, sqrt(2)/2 ).
In general: the x-values of the points of intersection of the circle x2 + y2 = 1 and y = mx can be found by
replacing the value of y in the equation of the circle with the value mx ---> x2 + (mx)2 = 1
---> x2 + m2x2 = 1 ---> x2(1 + m2) = 1 ---> x2 = 1 / (1 + m2)
---> lower value: x = - sqrt( 1 / (1 + m2) )
---> upper value: x = sqrt( 1 / (1 + m2) )
