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Radius of a circle touching lines \mathrm{y=\sqrt{2} x \: and \: \sqrt{2} y=x}, whose centre lies in the first quadrant and centre lies on a line \mathrm{x+y=3} is
 

Option: 1

\frac{3-\sqrt{3}}{2}

 


Option: 2

\frac{2-\sqrt{3}}{2}
 


Option: 3

\frac{\sqrt{6}-\sqrt{3}}{2}
 


Option: 4

\frac{1+\sqrt{3}}{2}


Answers (1)

best_answer

Centre lies on angle bisectors of intersecting tangents.

Angle bisector \mathrm{y=\sqrt{2} x \: and \: \sqrt{2} y=x \: are\: y= \pm x.}

As centre lies in the first quadrant so center lies on \mathrm{y=x}

Centre also lies on \mathrm{x+y=3}, so centre is \mathrm{\left(\frac{3}{2}, \frac{3}{2}\right).}

Radius = Perpendicular distance form \mathrm{\left(\frac{3}{2}, \frac{3}{2}\right) \text { to } \sqrt{2} x-y=0 }

Radius=\mathrm{\frac{\sqrt{2} \cdot\left(\frac{3}{2}\right)-\frac{3}{2}}{\sqrt{3}}=\frac{\sqrt{3}}{2}(\sqrt{2}-1)=\frac{\sqrt{6}-\sqrt{3}}{2}}

Hence option 3 is correct .

 

Posted by

Ritika Jonwal

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