An infinite number of tangents can be drawn f | vedclass

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(C) The equation of the circle is ${x^2} + {y^2} - 2x - 4y + \lambda = 0$.Comparing this with the general equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$

Vedclass · Accepted Answer

$5$

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(C) The equation of the circle is ${x^2} + {y^2} - 2x - 4y + \lambda = 0$.Comparing this with the general equation ${x^2} + {y^2} + 2gx + 2fy + c = 0$,we get $g = -1$,$f = -2$,and $c = \lambda$.The center of the circle is $(-g, -f) = (1, 2)$.For an infinite number of tangents to be drawn from a point to a circle,the point must lie on the circle,and the circle must be a point circle (radius $= 0$).Since the point $(1, 2)$ is the center,the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-2)^2 - \lambda} = \sqrt{1 + 4 - \lambda} = \sqrt{5 - \lambda}$.Setting the radius to $0$,we get $\sqrt{5 - \lambda} = 0$,which implies $5 - \lambda = 0$,so $\lambda = 5$.

An infinite number of tangents can be drawn from $(1, 2)$ to the circle ${x^2} + {y^2} - 2x - 4y + \lambda = 0$. Then,$\lambda = $

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