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Q.
The equation of the image of circle $x^2 + y^2 + 16x - 24y + 183 = 0$ by the line mirror $4x + 7y + 13 = 0$ is
Conic Sections
Solution:
The centre of the given circle is (-8, 12) and radius is 5.
The image of the circle will have the same radius, i.e. the radius of the required circle is 5.
The centre D of the required circle is the image of the centre C of the given circle in the line mirror. If D be $\left(\alpha, \, \beta\right)$ then
$\frac{\alpha + 8}{4} = \frac{\beta-12}{7} = -2\left[\frac{4 \times - 8 + 7\times 12 + 13}{4^{2}+7^{2}}\right]$ [See straight line]
Or, $\frac{\alpha + 8}{4} = \frac{\beta -12}{7} = \frac{-2\times65}{65} = -2$
$\therefore \alpha = -16, \,\beta = -2$
$\therefore $ Required circle is $\left(x+16\right)^{2} + \left(y+2\right)^{2} = 5^{2}$
