The angle between the tangents from a point on $x^{2} + y^{2} + 2 x + 4 y - 31 = 0$ to the circle $x^{2} + y^{2} + 2 x + 4 y - 4 = 0$ is

\frac{π}{6}

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\frac{π}{2}

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\frac{π}{4}

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\frac{π}{3}

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Solution

The correct option is D $\frac{π}{3}$
Given circles are concentric circles with centre of $(- 1, - 2)$
$γ_{1} = 6$ and $γ_{2} = 3$ are the radii of the circles
$s i n θ = \frac{3}{6} = \frac{1}{2}$
$θ = 30^{o}$
So, angle between the tangent from $s_{1} : x^{2} + y^{2} + 2 x + 4 y - 31 = 0$ to $s_{2} : x^{2} + y^{2} + 2 x + 4 y - 4 = 0$ is
$2 θ = 60^{o} = \frac{π}{3}$

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The angle between the tangents from a point on x2+y2+2x+4y−31=0 to the circle x2+y2+2x+4y−4=0 is

The correct option is D π3Given circles are concentric circles with centre of (−1,−2)γ1=6 and γ2=3 are the radii of the circlessinθ=36=12θ=30oSo, angle between the tangent from s1:x2+y2+2x+4y−31=0 to s2:x2+y2+2x+4y−4=0 is2θ=60o=π3

The angle between the tangents from a point on $x^{2} + y^{2} + 2 x + 4 y - 31 = 0$ to the circle $x^{2} + y^{2} + 2 x + 4 y - 4 = 0$ is