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Find the interval of the values of a for which the line x + y = 0 bisects two chords drawn from a point \mathrm{\left(\frac{1+\sqrt{2} a}{2}, \frac{1-\sqrt{2} a}{2}\right)to the circle \mathrm{2 x^2+2 y^2-(1+\sqrt{2} \cdot a) x-(1-\sqrt{2} \cdot a) y=0} 

Option: 1

a\: \epsilon (-\infty ,-2)\cup \left [ 2,\infty \right ] 


Option: 2

a\epsilon \left ( -\infty,2\right )


Option: 3

a\epsilon [-2,2]\cup [4,6]


Option: 4

a\epsilon [-2,\infty]


Answers (1)

best_answer

The given point lies on the given circle. The equation of the  chord  of the given circle, with (x1,y1) as its mid-point, is 

\mathrm{x x_1+y y_1-\frac{1+\sqrt{2} a}{4}\left(x+x_1\right)-\frac{1-\sqrt{2} a}{4}\left(y+y_1\right)=}\mathrm{x_1{ }^2+y_1{ }^2-\frac{1+\sqrt{2} a}{2} x_1-\frac{1-\sqrt{2} a}{2} y_1} 

Now (x1 , y1)  lies on the line \mathrm{x+y=0 \Rightarrow x_1+y_1=0 \Rightarrow y_1=-x_1}

Hence, \mathrm{2 x_1^2+\frac{3}{2} \sqrt{2} a x_1-\frac{1+2 a^2}{4}=0}

Since x_1 is real

]mathrm{\begin{array}{ll} 36 \times 2 a^2-32\left(1+2 a^2\right) \geq 0 & \\ \Rightarrow 9 a^2-4-8 a^2 \geq 0 & \Rightarrow a^2 \geq 4 . \\ \Rightarrow a \leq-2 \text { or } a \geq 2 & \Rightarrow a \in(-\infty,-2] \cup[2, \infty) . \end{array}}

 

 

 

 

 

Posted by

Divya Prakash Singh

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