  Question

# The circle x2+y2−4x−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√x2+y2=0, then the value of k is213−2

Solution

## The correct option is B 1Let OAB be the triangle in which the circle x2+y2−4x−4y+4=0 is inscribed. Assuming A=(a,0),B=(0,b), we get equation of line AB as xa+yb=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=a2,k=b2⋯(1) Given circle is  x2+y2−4x−4y+4=0 whose centre and radius is  C=(2,2),r=√22+22−4=2 Now, as line AB is tangent to the circle, distance from centre to tangent is equal to radius, so  ∣∣ ∣ ∣ ∣∣2a+2b−1√1a2+1b2∣∣ ∣ ∣ ∣∣=2 As origin and centre lies on the same side of the line AB, we get 2a+2b−1<0 So, 2a+2b−1√1a2+1b2=−2 ⇒2a+2b−ab+2√a2+b2=0 Using equation (1), we get ⇒4h+4k−4hk+2√4h2+4k2=0⇒h+k−hk+√h2+k2=0 So, the locus of the circumcenter(h,k) is x+y−xy+√x2+y2=0 Given locus is x+y−xy+k√x2+y2=0 On comparing both the locus, we get k=1  Suggest corrections   