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  • The equation(s) of circle which passes through the origin and cuts off chords of length b from lines y = x and y = -x is.Option: 1 <im

The equation(s) of circle which passes through the origin and cuts off chords of length b from lines y = x and y = -x is.

Option: 1

\mathrm{x^2+y^2+\sqrt{2} by =0}


Option: 2

\mathrm{x^2+y^2-\sqrt{2} by =0}


Option: 3

Both (a) and (b)


Option: 4

None of these.


Answers (1)

best_answer

The lines y = x and y = –x are at right angle to each other and the circle passes through origin

and \mathrm{\angle O A B=\pi / 2}
and hence AB is a diameter of length \mathrm{\sqrt{\left(b^2+b^2\right)}=b \sqrt{2}}.
\mathrm{ \therefore \quad r=\frac{b \sqrt{2}}{2}=\frac{b}{\sqrt{(2)}}=O C_1 . }
Clearly \mathrm{ C_1 } is centre on x-axis with co-ordinates \mathrm{ (b / \sqrt{2}, 0) }. Similarly the other centres could be
\mathrm{ C_2(-b / \sqrt{2}, 0) \text { or } C_3(0, b / \sqrt{2}) \text { or } C_4(0,-b / \sqrt{2}) \text {. } }
Hence the circles are \mathrm{ \left(x \pm \frac{b}{\sqrt{(2)}}\right)^2+y^2=\left(\frac{b}{\sqrt{(2)}}\right)^2 }
\mathrm{ \Rightarrow x^2+y^2 \pm \sqrt{2} b x=0 }
or \mathrm{ (x-0)^2+\left(y \pm \frac{b}{\sqrt{(2)}}\right)^2=\left(\frac{b}{\sqrt{(2)}}\right)^2 }

\mathrm{ \Rightarrow \quad x^2+y^2 \pm \sqrt{2} b y=0 . }

Posted by

Suraj Bhandari

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