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^…

Question

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 $r$ is equal to:

Accepted Answer

$$\begin{aligned}
& \frac{x+4}{1}=\frac{y-5}{2}=\frac{-2(4)}{5} \
& \Rightarrow \quad x=-4-\frac{8}{5}=-\frac{28}{5}, y=5-\frac{16}{5}=\frac{9}{5} \
& \therefore \quad \text { Image is }\left(\frac{-28}{5}, \frac{9}{5}\right)
\end{aligned}$$

Image lies on circle $(x+4)^2+(y-3)^2=r^2$

$$\begin{aligned}
& \left(\frac{-28}{5}+4\right)^2+\left(\frac{9}{5}-3\right)^2=r^2 \
& \Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \
& \Rightarrow r=2
\end{aligned}$$

Solutions · 1