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 circumcentre of the triangle is x+y−xy+k√x2+y2=0, then the value of k is equal to
A
2
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B
1
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C
3
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D
−2
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Solution
The correct option is B1 Let OAB be the triangle in which the circle x2+y2−4x−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, |2a+2b−ab|√a2+b2=2⇒|2(a+b)−ab|=2√a2+b2
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√a2+b2⇒2a+2b−ab+2√a2+b2=0⋯(1)
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=a2,k=b2⇒a=2h,b=2k Putting in equation (1), we get 4h+4k−4hk+4√h2+k2=0 Locus of the circumcentre is x+y−xy+√x2+y2=0