The equation of the image of the circle x^2 + | vedclass

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(D) The given circle is $x^2 + y^2 + 16x - 24y + 183 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 8, f = -12, c = 183$. The centre

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$x^2 + y^2 + 32x + 4y + 235 = 0$

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(D) The given circle is $x^2 + y^2 + 16x - 24y + 183 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 8, f = -12, c = 183$. The centre is $(-g, -f) = (-8, 12)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{64 + 144 - 183} = \sqrt{25} = 5$. Let the image of the centre $(-8, 12)$ in the line $4x + 7y + 13 = 0$ be $(h, k)$. Using the formula $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$: $\frac{h + 8}{4} = \frac{k - 12}{7} = -2 \frac{4(-8) + 7(12) + 13}{4^2 + 7^2} = -2 \frac{-32 + 84 + 13}{16 + 49} = -2 \frac{65}{65} = -2$. So,$h + 8 = -8 \implies h = -16$ and $k - 12 = -14 \implies k = -2$. The new centre is $(-16, -2)$ and the radius remains $5$. The equation of the image circle is $(x + 16)^2 + (y + 2)^2 = 5^2$. $x^2 + 32x + 256 + y^2 + 4y + 4 = 25$. $x^2 + y^2 + 32x + 4y + 235 = 0$.

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:

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