If the circle $S = x^{2} + y^{2} - 16 = 0$ intersects another circle $S' = 0$ of radius 5 in a such a manner that the common chord is of maximum length and has a slope equal to $\frac{3}{4}$ then the centre of $S' = 0$ can be

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$(\frac{9}{5}, \frac{- 12}{5})$

$(\frac{- 9}{5}, \frac{12}{5})$

$(\frac{9}{5}, \frac{12}{5})$

$(\frac{- 9}{5}, \frac{- 12}{5})$

answer is A, B.

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Detailed Solution

$x^{2} + y^{2} = 16 O A = 4 O (0, 0) = (x_{1}, y_{1}) A C = 5 \Rightarrow O C = 3 = | r |$
Slope of AB is 3/4 $\Rightarrow$ Slope of OC is $\tan θ = - 4 / 3$
$\Rightarrow \sin θ = 4 / 5, \cos θ = - 3 / 5$
Required centre $(x_{1} \pm r \cos θ, y_{1} \pm r \sin θ)$
i.e., $(- \frac{9}{5}, \frac{12}{5}) o r (\frac{9}{5}, - \frac{12}{5})$