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The equation of the image of the circle ${x^2} + {y^2} + 16x - 24y + 183 = 0$ by the line mirror 4x+7y+13=0 is:
A. ${x^2} + {y^2} + 32x - 4y + 235 = 0$
B. ${x^2} + {y^2} + 32x + 4y - 235 = 0$
C. ${x^2} + {y^2} + 32x - 4y - 235 = 0$
D. ${x^2} + {y^2} + 32x + 4y + 235 = 0$

Answer
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Hint: In this question, to find the image of the circle, we will find the image of the centre of the circle by the given line and radius will be the same. To find the image of a point ,we will write the equation of the line in symmetrical form and then find the image of the centre point of the circle.

Complete step by step answer:
The given equation of circle is ${x^2} + {y^2} + 16x - 24y + 183 = 0$ , we will convert this in standard form to find the centre of the circle.
The given circle equation can be written as:
${x^2} + 16x + 64 + {y^2} - 24y + 144 + - 25 = 0$
$ \Rightarrow {(x + 8)^2} + {(y - 12)^2} = 25 = {5^2}$ (1)
We know that the standard form of circle equation is given as:
$ \Rightarrow {(x - a)^2} + {(y - b)^2} = {r^2}$ (2)
Comparing equation 1 and 2, we have:
Centre: (-8 , 12) , radius = 5cm.
Image of the centre by line 4x+7y+13 =0 is given by:
$\dfrac{{x - ( - 8)}}{4} = \dfrac{{y - 12}}{7} = \dfrac{{(4( - 8) + 7(12) + 13 \times ( - 2))}}{{{4^2} + {7^2}}}$
On simplifying the above expression, we get:
$\dfrac{{x + 8}}{4} = \dfrac{{y - 12}}{7} = \dfrac{{65 \times - 2}}{{65}}$
$ \Rightarrow \dfrac{{x + 8}}{4} = - {\text{ and }}\dfrac{{y - 12}}{7} = - 2$
$ \Rightarrow $ x= -16 and y = -2.
Therefore, the new centre is ( -16,-2)
The radius is same as earlier i.e. r = 5cm
Therefore, the new equation of circle which is formed by taking the image of the given circle by the line 4x+7y+13=0 is given as:
$ \Rightarrow {(x + 16)^2} + {(y + 2)^2} = {5^2} = 25$
${x^2} + {y^2} + 32x + 4y + 235 = 0$
Hence, option D is the correct option.

Note:
In this type of question, you should know how to find the image of a point by a given line. The image of a point can also be found by assuming a point which lies on the given line and line joining the point and its image and then using the concept that point on the given line is mid point of given point and its image also the two lines are perpendicular.