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 circumcenter of the triangle is $x+y-xy+k\sqrt{x^2+y^2}=0$, then the value of $k$ is

• 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.
Assuming $A=(a,0),B=(0,b)$, we get equation of line $AB$ as
$\frac{x}{a}+\frac{y}{b}=1$

As $△OAB$ is right angle triangle, so its circumcenter is the mid point of $AB$,
Let the circumcentre be $P(h,k)$, so
$h=\frac{a}{2},k=\frac{b}{2}\cdots \left(1\right)$

Given circle is $x^2+y^2-4x-4y+4=0$ whose centre and radius is
$C=(2,2),r=\sqrt{2^2+2^2-4}=2$
Now, as line $AB$ is tangent to the circle, distance from centre to tangent is equal to radius, so
$\left|\frac{\frac{2}{a}+\frac{2}{b}-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right|=2$

As origin and centre lies on the same side of the line $AB$, we get
$\frac{2}{a}+\frac{2}{b}-1<0$
So,
$\frac{\frac{2}{a}+\frac{2}{b}-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=-2$
$\Rightarrow 2a+2b-ab+2\sqrt{a^2+b^2}=0$
Using equation $\left(1\right)$, we get
$\Rightarrow 4h+4k-4hk+2\sqrt{4h^2+4k^2}=0\Rightarrow h+k-hk+\sqrt{h^2+k^2}=0$
So, the locus of the circumcenter$(h,k)$ is
$x+y-xy+\sqrt{x^2+y^2}=0$
Given locus is $x+y-xy+k\sqrt{x^2+y^2}=0$
On comparing both the locus, we get
$k=1$