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circle_line_intersect_proof

circle_line_intersect_proof - Proof of Intersection of a...

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Proof of Intersection of a Line and a Circle P 1 ( x 1 , y 1 ) P 2 ( x 2 , y 2 ) r C ( x c , y c ) This problem is most easily solved if the circle is in implicit form: ( x - x c ) 2 + ( y - y c ) 2 - r 2 = 0 and the line is parametric: x = x 0 + ft y = y 0 + gt Substituting for (parametric line) x and y into the circle equation gives a quadratic equation in t : Two roots of which give points on the line where cuts the circle. t = f ( x c - x 0 ) + g ( y c - y 0 ) ± p r 2 ( f 2 + g 2 ) - ( f ( y c - y 0 ) - g ( x c - x 0 )) 2 ( f 2 + g 2 ) The roots maybe coincident if the line is tangential to the circle. P 1 ( x 1 , y 1 ) r C ( x c , y c ) If roots are imaginary then there is no intersection. 1
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Proof: The circle is in implicit form: ( x - x c ) 2 + ( y - y c ) 2 - r 2 = 0 (1) and the line is parametric: x = x 0 + ft y = y 0 + gt (2) Substituting for (parametric line) x and y from Eqn. 2 into the circle equation, Eqn. 1, gives a quadratic equation in t gives: ( x 0 + ft - x c ) 2 + ( y 0 + gt - y c ) 2 - r 2 = 0 (3) Let x d = x 0 - x c y d = y 0 - y c (4) Then we may write Eqn. 3 ( x d + ft ) 2 + ( y d + gt ) 2 - r 2 = 0 (5) Expanding the quadratic parts in Eqn. 5 gives: x 2 d + 2 fx d t + f 2 t 2 + y 2 d + 2
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