A conic passes through the point (2,4) and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then equation of auxiliary circle of the conic is
A
x2+y2=16
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B
x2+y2=25
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C
x2+y2=4√2
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D
none of these
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Solution
The correct option is Bx2+y2=16 2x=x−ym⇒ym=−x ⇒dyy+dxx=0⇒lnxy=c ⇒xy=c(rectangular hyperbola) x=2,y=4⇒equation is xy=8 The equation of auxiliary circle is x2+y2=2c2 ⇒x2+y2=16