if 4l^2 - 5m^2 + 6l +1=0 the prove that lx + my +1=0 line touches a fixed circle
if 4l^2 - 5m^2 + 6l +1=0 the prove that lx + my +1=0 line touches a fixed circle
abdul , 11 Years ago
Grade
Magical Mathematics[Interesting Approach]> cooordinate-geometry...
1 Answers
Amit Jain
Last Activity: 11 Years ago
Let the fixedcirclebe x^2 + y^2 + 2gx + 2fy + c = 0 such that line lx + my + 1 = 0 touches it under the condition 4l^2-5m^2+6l+1.
Length of perpendiculat from the centre (-g, -f) of the circle to the tangent line = radius, v(g^2 + f^2 - c)
=> (-lg - mf + 1)^2 = (l^2 + m^2)(g^2 + f^2 - c)
=> (f^2 - c)l^2 + (g^2 - c)m^2 - 2fglm + 2gl + 2fm - 1 = 0
Comparing with 4l^2-5m^2+6l+1,
(f^2 - c)/4 = (g^2 - c) / (-5) = g/3 = - 1 and f = 0
=> g = -3 and c = 4
=> the equation of the circle is
x^2 + y^2 - 6x + 4 = 0.
Length of perpendiculat from the centre (-g, -f) of the circle to the tangent line = radius, v(g^2 + f^2 - c)
=> (-lg - mf + 1)^2 = (l^2 + m^2)(g^2 + f^2 - c)
=> (f^2 - c)l^2 + (g^2 - c)m^2 - 2fglm + 2gl + 2fm - 1 = 0
Comparing with 4l^2-5m^2+6l+1,
(f^2 - c)/4 = (g^2 - c) / (-5) = g/3 = - 1 and f = 0
=> g = -3 and c = 4
=> the equation of the circle is
x^2 + y^2 - 6x + 4 = 0.
Amit Jain
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IIT Delhi
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