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Tangents PA and PB are drawn to circle\mathrm{(x-5)^2+(y-7)^2=1} from point P lying on  \mathrm{y=x+\frac{1}{x}}  Locus of circumcentre of triangle PAB is

Option: 1

\mathrm{4 x^2+4 x y-6 x+10 y-10=0}


Option: 2

\mathrm{4 x^2-4 x y-6 x+10 y-10=0}


Option: 3

\mathrm{4 x^2+4 x y+6 x+10 y-10=0}


Option: 4

none of these


Answers (1)

best_answer

Let\mathrm{P \equiv\left(t, t+\frac{1}{t}\right)} , now if centre of circle is (5, 7) ≡ C say then PACB is a cyclic quadrilateral, so circumcentre of triangle PAB is midpoint of PC i.e. 

\mathrm{\begin{aligned} & x=\frac{5+t}{2} \text { and } y=\frac{7+t+\frac{1}{t}}{2} \\ & \Rightarrow t=2 x-5 \Rightarrow(2 y-7)=(2 x-5)+\frac{1}{(2 x-5)} \\ & \Rightarrow 4 x^2-4 x y-6 x+10 y-10=0 \end{aligned}}

 

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