If the locus of the mid points of the chords of the circle $x^2+y^2=25$, which subtend a right angle at the…
If the locus of the mid points of the chords of the circle $x^2+y^2=25$, which subtend a right angle at the origin is given by $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$ then $|a|=$
$\frac{2}{5}$
$\frac{5}{\sqrt{2}}$
$\frac{2}{25}$
$5 \sqrt{2}$
Solution
$x^2+y^2=25 \Rightarrow r=5$
Let midpoint of chord $A B$ be $C$
$\sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}} \Rightarrow \mathrm{BC}=\frac{5}{\sqrt{2}}$
Let co-ordinates of C be $\left(x_1, y_1\right)$
By Pythagoras theorem
$\begin{aligned}
& \mathrm{OB}^2=\mathrm{OC}^2+\mathrm{BC}^2 \\
& \Rightarrow 25=x_1^2+y_1^2+\frac{25}{2} \Rightarrow \frac{25}{2}=x_1^2+y_1^2 \\
& \left.\Rightarrow \frac{x_1^2}{\frac{25}{2}}+\frac{y_1^2}{\frac{25}{2}}=1 \Rightarrow \alpha \right\rvert\,=\frac{5}{\sqrt{2}}
\end{aligned}$
