Let the lengths of intercepts on x- axis and y-axis made by the circle x2+y2+ax+2ay+c=0, (a<0) be 22 and 25, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x+2y=0, is equal to :

x2+y2+ax+2ay+c=0Centre of circle is (−a2,−a) and radius 5a24−c=6So distance of tangent to the circle from origin |0+0+λ5|=6

Let the lengths of intercepts on x- axis and y-axis made by the circle x2+y2+ax+2ay+c=0, (a<0) be 22 and 25, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x+2y=0, is equal to :

x2+y2+ax+2ay+c=0Centre of circle is (−a2,−a) and radius 5a24−c=6So distance of tangent to the circle from origin |0+0+λ5|=6

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Let the lengths of intercepts on x- axis and y-axis made by the circle $x^{2} + y^{2} + a x + 2 a y + c = 0$ , $(a < 0)$ be $2 \sqrt{2}$ and $2 \sqrt{5}$ , respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x + 2 y = 0$ , is equal to :

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$\sqrt{6}$

$\sqrt{10}$

$\sqrt{11}$

$\sqrt{7}$

answer is A.

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Detailed Solution

$x^{2} + y^{2} + a x + 2 a y + c = 0$
Centre of circle is $(\frac{- a}{2}, - a)$ and radius $\sqrt{\frac{5 a^{2}}{4} - c} = \sqrt{6}$
So distance of tangent to the circle from origin $| \frac{0 + 0 + λ}{\sqrt{5}} | = \sqrt{6}$

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Let the lengths of intercepts on x- axis and y-axis made by the circle x2+y2+ax+2ay+c=0, (a<0) be 22 and 25, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x+2y=0, is equal to :

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