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

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(D) The centre of the given circle $x^2 + y^2 + 16x - 24y + 183 = 0$ is $C(-8, 12)$ and its radius $r = \sqrt{(-8)^2 + (12)^2 - 183} = \sqrt{64 + 144

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

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(D) The centre of the given circle $x^2 + y^2 + 16x - 24y + 183 = 0$ is $C(-8, 12)$ and its radius $r = \sqrt{(-8)^2 + (12)^2 - 183} = \sqrt{64 + 144 - 183} = \sqrt{25} = 5$.The image of the circle under the line mirror $4x + 7y + 13 = 0$ will have the same radius $r = 5$,but its centre will be the reflection of $C(-8, 12)$ across the line.Let the image centre be $C'(h, k)$. The midpoint of $CC'$ is $\left(\frac{h-8}{2}, \frac{k+12}{2}ight)$,which lies on the line $4x + 7y + 13 = 0$:$4\left(\frac{h-8}{2}ight) + 7\left(\frac{k+12}{2}ight) + 13 = 0 \implies 2h - 16 + 3.5k + 42 + 13 = 0 \implies 4h + 7k + 78 = 0$ $(i)$.The slope of $CC'$ is $\frac{k-12}{h+8}$,which must be perpendicular to the line $4x + 7y + 13 = 0$ (slope $-4/7$):$\left(\frac{k-12}{h+8}ight) 	imes \left(-\frac{4}{7}ight) = -1 \implies 7(k-12) = 4(h+8) \implies 7k - 84 = 4h + 32 \implies 4h - 7k + 116 = 0$ (ii).Adding $(i)$ and (ii): $8h + 194 = 0 \implies h = -16$. Substituting into $(i)$: $4(-16) + 7k + 78 = 0 \implies -64 + 7k + 78 = 0 \implies 7k = -14 \implies k = -2$.The new centre is $(-16, -2)$ and radius is $5$. The equation is $(x+16)^2 + (y+2)^2 = 5^2$,which simplifies to $x^2 + 32x + 256 + y^2 + 4y + 4 = 25$,or $x^2 + y^2 + 32x + 4y + 235 = 0$.

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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