Question 665773
x^2+y^3=13 

The co ordinates of center are (0,0)
point on circle = (-2,3)

The slope of radius = (0-3)/(0-(-2))

= -3/2

The tangent is always perpendicular to the radius.
so slope of tangent will be 2/3 ( negative reciprocal)
the tangent passes through (-2,3)
m=		 2/3						 
Y	=	m	x	+	b			
3.00	=	 2/3	*	-2	+	b		
3.00	=	-1 1/3	+	b				
b=	3    	+	1 1/3					
b=	13/ 3							
So the equation  will be								
Y 	=	 2/3	x	+	13/3			
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