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The angle between the circles $x^{2} + y^{2} + 4 x - 14 y + 28 = 0$ and $x^{2} + y^{2} - 12 x - 6 y - 4 = 0$ is

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$45^{\circ}$

$60^{\circ}$

${Cos}^{- 1} (3 / 35)$

$\cos^{- 1} (2 / \sqrt{5})$

answer is B.

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Detailed Solution

For the given circles we have $C_{1} = (- 2, 7), r_{1} = \sqrt{4 + 49 - 28} = 5$ ;

$C_{2} = (6.3), r_{2} = \sqrt{36 + 9 + 4} = 7, and d = C_{1} C_{2} = \sqrt{64 + 16} = \sqrt{80}$

Let $θ$ be the angle between the circles

$\cos θ = \frac{d^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}} = \frac{80 - 25 - 49}{2 \cdot 5 \cdot 7} = \frac{6}{2 \cdot 5 \cdot 7} = \frac{3}{35} . ∴ θ = {Cos}^{- 1} (\frac{3}{35})$

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