KCET · Maths · Circle
The locus of the mid points of the chords of the circle \(x^{2}+y^{2}=4\) which subtend a right angle at the origin is
- A \(x^{2}+y^{2}=1\)
- B \(\mathrm{x}^{2}+\mathrm{y}^{2}=2\)
- C \(x+y=1\)
- D \(x+y=2\)
Answer & Solution
Correct Answer
(B) \(\mathrm{x}^{2}+\mathrm{y}^{2}=2\)
Step-by-step Solution
Detailed explanation
Let mid point of the chord \(\mathrm{AB}\) is \(\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\).
In \(\Delta \mathrm{COB}, \sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}\)
\[
\begin{array}{ll}
\Rightarrow & \frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2} \\
\Rightarrow & \mathrm{BC}=\sqrt{2}
\end{array}
\]
Using Pythagoras theorem
\(\mathrm{OB}^{2}=\mathrm{OC}^{2}+\mathrm{CB}^{2}\)
\(\Rightarrow \quad(2)^{2}=\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+(\sqrt{2})^{2}\)
\(\Rightarrow \quad \mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}=2\)
Hence, locus of mid point of chord is
\[
x^{2}+y^{2}=2
\]
In \(\Delta \mathrm{COB}, \sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}\)
\[
\begin{array}{ll}
\Rightarrow & \frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2} \\
\Rightarrow & \mathrm{BC}=\sqrt{2}
\end{array}
\]
Using Pythagoras theorem
\(\mathrm{OB}^{2}=\mathrm{OC}^{2}+\mathrm{CB}^{2}\)
\(\Rightarrow \quad(2)^{2}=\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+(\sqrt{2})^{2}\)
\(\Rightarrow \quad \mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}=2\)
Hence, locus of mid point of chord is
\[
x^{2}+y^{2}=2
\]
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