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:
Solution
<p>$$\begin{aligned}
& \frac{x+4}{1}=\frac{y-5}{2}=\frac{-2(4)}{5} \\
& \Rightarrow \quad x=-4-\frac{8}{5}=-\frac{28}{5}, y=5-\frac{16}{5}=\frac{9}{5} \\
& \therefore \quad \text { Image is }\left(\frac{-28}{5}, \frac{9}{5}\right)
\end{aligned}$$</p>
<p>Image lies on circle $(x+4)^2+(y-3)^2=r^2$</p>
<p>$$\begin{aligned}
& \left(\frac{-28}{5}+4\right)^2+\left(\frac{9}{5}-3\right)^2=r^2 \\
& \Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \\
& \Rightarrow r=2
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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