If the circles are given by $x^{2} + y^{2} + 8 x - 2 y - 9 = 0 a n d x^{2} + y^{2} - 2 x + 8 y - 7 = 0$ , then the angle at which these circles intersect is

\frac{2 π}{3}

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\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{2}$
$x^{2} + y^{2} + 8 x - 2 y - 9 = 0$
$C_{1} = (- 4, 1)$ and $r_{1} = \sqrt{4^{2} + 1^{2} + 9} = \sqrt{26}$

$x^{2} + y^{2} - 2 x + 8 y - 7 = 0$
$C_{2} \equiv (1, - 4)$ and $r_{2} = \sqrt{1^{2} + 4^{2} + 7} = \sqrt{24}$
$∴ cos (180 - θ) = \frac{r {_{1}}^{2} + r {_{2}}^{2} - (c_{1} c_{2})^{2}}{2 r_{1} r_{2}}$
$\Rightarrow - cos θ = \frac{24 + 26 - 50}{2 \sqrt{24} \sqrt{26}}$
$\Rightarrow cos θ = 0$
$\Rightarrow θ = 90^{0} = \frac{π}{2}$ is the angle between the circles
Hence, option D.

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