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Question

The tangent at the point $$P$$ on the rectangular hyperbola $$xy={k}^{2}$$ with $$C$$ intersects the coordinate axes at $$Q$$ and $$R$$ . Locus of the circumcentre of triangle $$CQR$$ is


A
x2+y2=2k2
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B
x2+y2=k2
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C
xy=k2
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D
none of these
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Solution

The correct option is C $$xy={k}^{2}$$
Given:A tangent at the point $$P$$ on the rectangular hyperbola $$xy={k}^{2}$$ with $$C$$ intersects the coordinate axes at $$Q$$ and $$R$$ where $$C\left(0,0\right)$$ is the center of hyperbola.
$$\therefore\,\triangle{CQR}$$ is a rightangled triangle where $$\angle{C}={90}^{\circ}$$
The circumcentre of a right angled triangle is the mid-point of its hypotenuse.
Coordinates of hyperbola $$xy={k}^{2}$$ is $$\left(kt,\dfrac{k}{t}\right)$$
The tangent at $$\left(kt,\dfrac{k}{t}\right)$$ is $$x+y{t}^{2}=2kt$$
cuts the $$y-$$ axis at $$Q$$
At $$x=0\Rightarrow\,y{t}^{2}=2kt$$
$$\Rightarrow\,y=\dfrac{2kt}{{t}^{2}}=\dfrac{2k}{t}$$
$$\therefore\,$$co-ordinates of $$Q$$ are $$\left(0,\dfrac{2k}{t}\right)$$
At $$x-$$axis we have $$y=0$$ at $$R$$
$$\therefore\,x+0=2kt$$
$$\Rightarrow\,x=2kt$$
Hence coordinates of $$R$$ is $$\left(2kt,0\right)$$
Midpoint of $$QR$$ is $$\left(\dfrac{0+2kt}{2},\dfrac{\dfrac{2k}{t}+0}{2}\right)=\left(kt,\dfrac{k}{t}\right)$$
$$\therefore\,x=kt,\,y=\dfrac{k}{t}$$
$$\Rightarrow\,xy=kt\times\dfrac{k}{t}={k}^{2}$$
Hence the locus of the circumcentre of the triangle $$CQR$$ is $$xy={k}^{2}$$

1471304_311860_ans_87f01ac1324248e589abcf511be2815f.PNG

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