If the image of the point $(-4,5)$ in the line $x+2 y=2$ lies on the circle $(x+4)^2+(y-3)^2=r^2$, then $\mathrm{r}$ is equal to:

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Aicharya · Accepted Answer

Image of point $(-4,5)$ $\frac{x-x_1}{a}=\frac{y-y_1}{b}=-2\left(\frac{a x_1+b y_1+c}{a^2+b^2}ight)$ Line: $x+2 y-2=0$ $\frac{x+4}{1}=\frac{y-5}{2}=-2\left(\frac{-4+10-2}{1^2+2^2}ight)$ $=\frac{-8}{5}$ $x=-4-\frac{8}{5}=-\frac{28}{5}$ $y=-\frac{16}{5}+5=\frac{9}{5}$ Point lies on circle $(x+4)^2+(y-3)^2=r^2$ $\begin{aligned} & \frac{64}{25}+\left(\frac{9}{5}-3ight)^2=r^2 \ & \frac{100}{25}=r^2, r=2\end{aligned}$

If the image of the point $(-4,5)$ in the line $x+2 y=2$ lies on the circle $(x+4)^2+(y-3)^2=r^2$, then…

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