The image of the circle $x^{2} + y^{2} + 16 x - 24 y + 183 = 0$ in the line mirror
$4 x + 7 y + 13 = 0,$ is

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$(x + 16)^{2} + (y + 2)^{2} = 5^{2}$

$(x + 16)^{2} + (y + 2)^{2} = 5^{2}$ .

$(x + 16)^{2} + (y - 2)^{2} = 5^{2}$

$(x - 16)^{2} + (y - 2)^{2} = 5^{2}$

answer is A.

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Detailed Solution

The equation of the given circle is

$\begin{array}{l} x^{2} + y^{2} + 16 x - 24 y + 183 = 0 \\ \Rightarrow (x^{2} + 16 x + 64) + (y^{2} - 24 y + 144) = 25 \\ \Rightarrow (x + 8)^{2} + (y - 12)^{2} = 5^{2} \\ \Rightarrow {x - (- 8)}^{2} + (y - 12)^{2} = 5^{2} \end{array}$

Clearly, the centre of this circle is (- 8, 12) and radius= 5.

The image of the circle (i) in the line mirror $4 x + 7 y + 13 = 0$ is a

ircle whose centre is the image of the point ( - 8, 12) in the line

mirror $4 x + 7 y + 13 = 0$ and radius same as that of the given circle.

Let $(h, k)$ be the image of (- 8, 12) in the line mirror

$4 x + 7 y + 13 = 0 .$ The,

$\begin{array}{l} \frac{h + 8}{4} = \frac{k - 12}{7} = \frac{- 2 (4 \times (- 8) + 7 \times 12 + 13)}{(4^{2} + 7^{2})} \\ \Rightarrow \frac{h + 8}{4} = \frac{k - 12}{7} = \frac{- 130}{65} \Rightarrow h = - 16, k = - 2 \end{array}$