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

Option: 1

-1


Option: 2

1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

Let OAB be the  triangle which has two sides OA  and OB along the axes. Let the equation of the third side AB be \mathrm{\frac{x}{a}+\frac{y}{b}=1}

∴ coordinate of vertices are O(0, 0), A(a, 0) and  B(0, b) 

The circle inscribed in the ΔOAB is

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

whose centre is (2, 2) and radius = 2. 

Since (2, 2) is the incentre of ΔOAB, we have


\mathrm{\frac{0 \cdot \sqrt{a^2+b^2}+a \cdot b+0 \cdot a}{a+b+\sqrt{a^2+b^2}}=2} and

\mathrm{\frac{0 \cdot \sqrt{a^2+b^2}+b \cdot a+0 \cdot b}{a+b+\sqrt{a^2+b^2}}=2}

\mathrm{\therefore a b=2\left[a+b+\sqrt{a^2+b^2}\right]}-----(1)

If P(α, β) is the circum-centre of ΔOAB, then 

\mathrm{a=\frac{a}{2}} and \mathrm{\beta =\frac{b}{2}} so (1) gives

\mathrm{4 \alpha \beta=2\left[2 \alpha+2 \beta+2 \sqrt{\alpha^2+\beta^2}\right]}

∴ locus of P(α, β) is

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

But it is given that  locus of circumcentre P(α, β) is 

\mathrm{x+y-x y+k \sqrt{x^2+y^2}=0} -------(3)

Comparing (2) and (3), we get  k = 1.

 

 

 

 

 

 

 

 

Posted by

Sanket Gandhi

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