The common chord of the circles $x^{2} + y^{2} - 4 x - 4 y = 0$ and $2 x^{2} + 2 y^{2} = 32$ subtends at the origin an angle equal to

\frac{π}{3}

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\frac{π}{4}

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\frac{π}{6}

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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 $C_{1} - C_{2} = 0 \Rightarrow x^{2} + y^{2} - 16 - (x^{2} + y^{2} - 4 x - 4 y) \Rightarrow x + y = 4$
common chord and circle intersect each other at $(0, 4) & (4, 0)$
Therefore it will form an angle $90 °$ at $(0, 0)$

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Similar questions

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

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

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

C

\sqrt{10}

y = 2 x + \frac{5}{\sqrt{2}}

45^{\circ}

C

x^{2} + (y - λ)^{2} = 16

x^{2} + y^{2} = 16

λ

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

t a n^{- 1} (\frac{7}{4})

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