The common chord of the circles $x^{2} + y^{2} - 4 x - 4 y = 0$ and $x^{2} + y^{2} = 16$ subtends at the origin an angle equal to

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{2}$
Equation of common chord of circles $s_{1}$ and $s_{2}$ is given by
$s_{1} - s_{2} = 0$
$∴ e q^{n}$ of common chord is
$4 x + 4 y = 16$
$x + y = 4$ ----(2)
So, angle subtended by $x + y = 4$ at origin is $\frac{π}{2}$

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Similar questions

The common chord of $x^{2} + y^{2} - 4 x - 4 y = 0$ and $x^{2} + y^{2} = 16$ subtends at the origin an angle equal to

x^{2} + y^{2} - 4 x - 4 y = 0

2 x^{2} + 2 y^{2} = 32

x^{2} + y^{2} = 4

\frac{π}{2}

x^{2} + y^{2} = 1.

x^{2} + y^{2} = r^{2}

^{'} θ^{'}

x^{2} + y^{2} = r^{2} {cos}^{2} (\frac{θ}{2})

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

π / 3

x^{2} + y^{2} = 41

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

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