Question 203944: Find m such that the line y=mx+3 is tangent to the circle x^2+y^2=2
Could you guide me how can I find m in a circle equation?thanks
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! The equation of a circle is x^2 +y^2 =r^2,,,,in this case x^2+y^2=2,,,
or r=sqrt2
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The tan line has the eqn,,,,y=mx+3,,,,with std eqn,,y=mx=b,,,,b=3
or the y intercept is at (0,3)
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Drawing a rough sketch helps. Circle about origin with r= sqrt2 = 1.414
tan line thru (0,3) to either side of circle.
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Remember that a line tan to a circle, touches only one point, perpendicular to the radius.
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Adding such a line to the sketch, finds a rt triangle, one vertex at (0,0), another at (0,3) .
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Legs of this triangle are 3 and sqrt2.
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let angle formed by vertical axis and tan line be "a".
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sin a = sqrt2 /3 = .4714,,,,,a=28.1255 degrees
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tan a = x/y of line tan to circle,,,or y/x= 1/tan a=1/tan 28.1255= 1.87
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but m=y/x =1.87,,,,this is negative on rt side of circle and positive on left side.
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from,,y=mx +b, lt side ln is y=1.87x +3,,rt side ln is y= -1.87x +3
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