Q.202. Circle and tangent to a circle in coordinate geometry

Question 202.
If 4l^2 - 5m^2 + 6l + 1 = 0, prove that the line lx + my + 1 = 0 touches a fixed circle.

Answer 202.
Let the fixed circle be 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 = 0.

Length of perpendiculat from the centre (-g, -f) of the circle to the tangent line = radius, √(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 which being uniques is the fixed circle.

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