CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The circle x2+y24x4y+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+yxy+kx2+y2=0, then the value of k is
  1. 2
  2. 1
  3. 3
  4. 2


Solution

The correct option is B 1
Let OAB be the triangle in which the circle x2+y24x4y+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+y24x4y+4=0 whose centre and radius is 
C=(2,2),r=22+224=2
Now, as line AB is tangent to the circle, distance from centre to tangent is equal to radius, so 
∣ ∣ ∣ ∣2a+2b11a2+1b2∣ ∣ ∣ ∣=2

As origin and centre lies on the same side of the line AB, we get
2a+2b1<0
So,
2a+2b11a2+1b2=2
2a+2bab+2a2+b2=0
Using equation (1), we get
4h+4k4hk+24h2+4k2=0h+khk+h2+k2=0
So, the locus of the circumcenter(h,k) is
x+yxy+x2+y2=0
Given locus is x+yxy+kx2+y2=0
On comparing both the locus, we get
k=1

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image