Question

The circle $x^2+y^2-4x-4y+4=0$ is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is $x+y-xy+k\sqrt{x^2+y^2}=0$, then the value of $k$ is equal to

• A. $2$
• B. $1$
• C. $3$
• D. $-2$

Answer 1

The correct option is B $1$

Let $OAB$ be the triangle in which the circle $x^2+y^2-4x-4y+4=0$ is inscribed.
Let points $A=(a,0)$ and $B=(0,b)$
So, the equation of the line $AB$ is
$bx+ay-ab=0$


Distance from the centre $(2,2)$ to the line $AB$ is equal to radius,
$\frac{|2a+2b-ab|}{\sqrt{a^2+b^2}}=2\Rightarrow |2(a+b)-ab|=2\sqrt{a^2+b^2}$

As origin and $C(2,2)$ lies on the same side of line $AB$, so
$2(a+b)-ab<0$
So,
$-2(a+b)+ab=2\sqrt{a^2+b^2}\Rightarrow 2a+2b-ab+2\sqrt{a^2+b^2}=0\cdots \left(1\right)$

Let $P(h,k)$ be the circumcentre of $△OAB$
As $△OAB$ is a right angled triangle at $O$, so circumcentre is the mid point of line segment $AB$
$h=\frac{a}{2},k=\frac{b}{2}\Rightarrow a=2h,b=2k$
Putting in equation $\left(1\right)$, we get
$4h+4k-4hk+4\sqrt{h^2+k^2}=0$
Locus of the circumcentre is
$x+y-xy+\sqrt{x^2+y^2}=0$

Hence, $k=1$