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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+k\left(x^2+y^2\right)^{\frac{1}{2}} =0} ,

then value of |k| is

Option: 1

4


Option: 2

3


Option: 3

2


Option: 4

1


Answers (1)

\mathrm{x^2+y^2-4 x-4 y+4=0}
Centre (2,2), radius r=2
Let the equation of the third side is
\mathrm{ \frac{x}{p}+\frac{y}{q}=1 \text { or } q x+p y-p q=0 }.................(i)
Since  \mathrm{ \angle P O Q }  is a right angle, PQ is a diameter of the circle passing through O,P,Q.
Its centre R(h, k) is midpoint of PQ

\mathrm{ \therefore h=\frac{p}{2}, k=\frac{q}{2} }
Using perpendicular distance formula, we have
\mathrm{ \left|\frac{q(2)+p(2)-p q}{\sqrt{p^2+q^2}}\right|=2 }

\mathrm{ \mathrm{ \Rightarrow\left|\frac{4 k+4 h-4 h k}{\sqrt{4 h^2+4 k^2}}\right|=2 }

\mathrm{ \Rightarrow h+k-h k= \pm \sqrt{h^2+k^2} \\ }\mathrm{ \therefore \text { Locus of } R(h, k) \text { is } \\ }

\mathrm{ x+y-x y \pm \sqrt{x^2+y^2}=0 }
But the locus of circumcentre is given as
\mathrm{ x+y-x y+k \sqrt{x^2+y^2}=0 . \\ }
\mathrm{ \therefore \quad k= \pm 1 . }
 

Posted by

Ramraj Saini

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