(d) The centre of the required circle is the image of the centre $( – 8,\;12)$ with respect to the line mirror $4x + 7y + 13 = 0$ and radius is equal to the radius of the given circle.
Let $(h, k)$ be the image of the point $( – 8,\;12)$ with respect to the line mirror. Then the mid-point of the line joining $C( – 8,\;12)$ and $P(h,\;k)$ lies on the line mirror.
$\therefore \;4\left( {\frac{{h – 8}}{2}} \right) + 7\left( {\frac{{k + 12}}{2}} \right) + 13 = 0$
or $4h + 7k + 78 = 0$….(i)
Also $CP$ is perpendicular to $4x + 7y + 13 = 0$
$\therefore $ $\frac{{k – 12}}{{h + 8}} \times – \frac{4}{7} = – 1$ or $7h – 4k + 104 = 0$….(ii)
Solving (i) and (ii), $h = – 16,\;k = – 2$.
Thus the centre of the image circle is $( – 16,\; – 2)$. The radius of the image circle is same as the radius of ${x^2} + {y^2} + 16x – 24y + 183 = 0$ i.e., $5$.
Hence the equation of the required circle is
${(x + 16)^2} + {(y + 2)^2} = {5^2}$ i.e. ${x^2} + {y^2} + 32x + 4y + 235 = 0$.