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Unformatted text preview: Proof of an implicit line equation through a point (not on the circle) that is tangent to a general circle Derivation of an equation that calculates the the general equation of an implicit line through a point (not on the circle) that is tangent to a general circle: P ( x, y ) J ( x j , y j ) r C ( x c , y c ) 1 We wish to find the implicit equation of the tangent ax + by + c = 0 The coefficients a and b are obtained from: a = r ( x c- x j )- ( y c- y j ) p ( x c- x j ) 2 + ( y c- y j ) 2- r 2 ( x c- x j ) 2 + ( y c- y j ) 2 b = r ( y c- y j ) + ( x c- x j ) p ( x c- x j ) 2 + ( y c- y j ) 2- r 2 ( x c- x j ) 2 + ( y c- y j ) 2 (1) c then obtained from the fact that the tangent passes through J : c =- ax j- by j See http://www.netsoc.tcd.ie/ jgilbert/maths site/applets/circles/tangents to circles.html for a java applet demo. Proof: 1 P ( x, y ) J ( x j , y j ) r C ( x c , y c ) 1 We know the coordinates of the circle ( x c ,y c ) and its radius, r ....
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This note was uploaded on 01/24/2012 for the course CM 0268 taught by Professor Davidmarshall during the Winter '11 term at Cardiff University.
- Winter '11