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標題:
Non-Linear Coordinate Geometry (AS)
發問:
the line joning the pointA(0,5)andB(4,1)is a tangent to circle whose centre,C,is at the point(5,4) A,find the equation of the lineAB b,find the equation of the line through Cwhich is perpendicular to AB c,find the coordinates of the point of contact of the line AB with the circle d,find the equation of the circle
最佳解答:
(a) use two-point form the equation of the line AB (y-5)/(x-0)=(1-5)/(4-0) y-5=-x x+y-5=0 (b) let the equation of the line through C which is perpendicular to AB is L the slope of AB is -1 so the slope of L is 1 the equation of L is y-4=x-5 x-y-1=0 (c) The intersection of the equation AB and L is the coordinates of the point of contact of the line AB with the circle AB:x+y-5=0 L:x-y-1=0 from L, x=y+1 sub into AB y+1+y-5=0 y=2 x=3 the coordinates of the point is (3,2) (d) the equation of the circle is (x-5)^2+(y-4)^2=(5-3)^2+(4-2)^2 x^2+y^2-10x-8y+25+8=0 2006-11-19 03:22:35 補充： 少了最後個行x^2 y^2-10x-8y 33=0
其他解答:
Non-Linear Coordinate Geometry (AS)
發問:
the line joning the pointA(0,5)andB(4,1)is a tangent to circle whose centre,C,is at the point(5,4) A,find the equation of the lineAB b,find the equation of the line through Cwhich is perpendicular to AB c,find the coordinates of the point of contact of the line AB with the circle d,find the equation of the circle
最佳解答:
(a) use two-point form the equation of the line AB (y-5)/(x-0)=(1-5)/(4-0) y-5=-x x+y-5=0 (b) let the equation of the line through C which is perpendicular to AB is L the slope of AB is -1 so the slope of L is 1 the equation of L is y-4=x-5 x-y-1=0 (c) The intersection of the equation AB and L is the coordinates of the point of contact of the line AB with the circle AB:x+y-5=0 L:x-y-1=0 from L, x=y+1 sub into AB y+1+y-5=0 y=2 x=3 the coordinates of the point is (3,2) (d) the equation of the circle is (x-5)^2+(y-4)^2=(5-3)^2+(4-2)^2 x^2+y^2-10x-8y+25+8=0 2006-11-19 03:22:35 補充： 少了最後個行x^2 y^2-10x-8y 33=0
其他解答:
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