See only find the center of the imaged circle coz the rest will be same

Equation of the Image of a Circle

To find the equation of the image of a circle by a line mirror, we need to follow a step-by-step process. In this case, we are given the equation of the circle and the equation of the line mirror. Let's proceed with the solution.

Step 1: Write the equation of the given circle
The equation of the given circle is x^2 + y^2 + 16x - 24y + 183 = 0.

Step 2: Find the center and radius of the given circle
To find the center and radius of the circle, we need to rewrite the equation in the standard form by completing the square.

x^2 + 16x + y^2 - 24y + 183 = 0
(x^2 + 16x + 64) + (y^2 - 24y + 144) = 64 + 144 - 183
(x + 8)^2 + (y - 12)^2 = 5

From the standard equation form, we can determine that the center of the circle is (-8, 12) and the radius is the square root of 5.

Step 3: Find the equation of the line perpendicular to the given line mirror
To find the equation of the line perpendicular to the given line mirror, we need to determine the slope of the line mirror. The given line mirror equation is 4x + 7y + 13 = 0.

The slope of the line mirror is given by -4/7. Since the line perpendicular to it will have a negative reciprocal slope, the slope of the perpendicular line is 7/4.

Step 4: Find the equation of the perpendicular line passing through the center of the circle
Using the slope-intercept form of the equation of a line (y = mx + c), we can substitute the center point (-8, 12) and the slope of the perpendicular line (7/4) to find the equation.

y - 12 = (7/4)(x + 8)
4y - 48 = 7x + 56
7x - 4y + 104 = 0

Step 5: Find the equation of the image of the circle
To find the equation of the image of the circle, we need to find the equations of the tangents from the center of the circle to the perpendicular line. These tangents will be perpendicular to the line mirror.

The equation of a tangent from a point (h, k) to a line Ax + By + C = 0 is given by:
(2Ah + Bk + C)/(A^2 + B^2) = 0

Using this formula, we can find the equations of the tangents from (-8, 12) to the line 7x - 4y + 104 = 0.

Tangent 1:
(2(7)(-8) + (-4)(12) + 104)/(7^2 + (-4)^2) = 0
(-56 - 48 + 104)/(49 + 16) = 0
0/65 = 0

Tangent 2:
(2(7)(-8) +