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The circle \mathrm{x^2+y^2-4 x-4 y+4=0} is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is \mathrm{x+y-x y+\mathrm{k}\left(\mathrm{x}^2+\mathrm{y}^2\right)^{1 / 2}=0}. Find \mathrm{k}.
 

Option: 1

1


Option: 2

2


Option: 3

\frac{1}{2}


Option: 4

\frac{3}{2}


Answers (1)

best_answer

The given circle is \mathrm{x^2+y^2-4 x-4 y=0}. Thus can be re-written as

\mathrm{(x-2)^2+(y-2)^2=4} which has center \mathrm{C}(2,2) and radius 2 .
Let the equation of third side is \mathrm{\frac{x}{a}+\frac{y}{b}=1} (equation of \mathrm{AB} )

Length of perpendicular from (2,2) on \mathrm{AB}=$ radius $=\mathrm{CM}

\therefore \frac{\left|\frac{2}{a}+\frac{2}{b}-1\right|}{\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}}=2

Since origin and (2,2) lie on the same side of \mathrm{A B}

\therefore \quad-\frac{\left(\frac{2}{a}+\frac{2}{b}-1\right)}{\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}}=2
or \quad \frac{2}{a}+\frac{2}{b}-1=-2 \sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}.................(i)

Since \angle \mathrm{AOB}=\pi / 2

Hence \mathrm{AB} is the diameter of the circle passing through \triangle \mathrm{OAB},mid point of \mathrm{AB} is the center of the circle i.e., \left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)
Let center be \mathrm{(h, k) \equiv\left(\frac{a}{2}, \frac{b}{2}\right) then \: a=2 h \& b=2 h}

Substituting the values of a and b in (i) then

\mathrm{\frac{2}{2h}+\frac{2}{2k}-1=-2\sqrt{\frac{1}{4h^{2}}+\frac{1}{4k^{2}}}}

\mathrm{\Rightarrow \frac{1}{h}+\frac{1}{k}-1=-\left(\sqrt{\frac{1}{h^2}+\frac{1}{k^2}}\right) }
or \mathrm{ h+k-h k+\sqrt{\left(h^2+k^2\right)}=0 }

\mathrm{\therefore \quad \text { Locus of } M(h, k) \text { is } }

\mathrm{ x+y-x y+\sqrt{\left(x^2+y^2\right)}=0}

Hence the required value of \mathrm{\mathrm{k}} is 1 .

Hence option 1 is correct.

Posted by

Riya

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