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Equation of tangent of the circle x*2+y*2-8x-2y+12=0 at the points whose ordinates are 1 |
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Answer» Kindly mention the absicsa too Let the point on the circle be (x,1)Where the tangent is also passing from the same point. x2+(1)2-8x-2(1)+12=0 x2-8x+11=0 Where x=-b+-√b2-4ac /2 x=4+-√5 The equation of the tangents is x=4+√5 and x=4-√5 |
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