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the equation of a circle,which passes through the points(2,0),(0,2) and has a smallest possible radius is ?

the equation of a circle,which passes through the points(2,0),(0,2) and has a smallest possible radius is ?

Grade:Upto college level

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
let the equation of required circle is x2+y2+2gx+2fy+c=0............(1)
As 1 passes through (2,0) and (0,2)
we get 4+4g+c=0 and 4+4f+c=0
so g=f=-(4+c)/2
Let r be the radius of circle (1) then
r=sqrt(g2+f2-c)
=sqrt(((4+c)/2)2+((4+c)/2)2-c)
=sqrt((c2+6c+16)/2)
so r2=(c2+6c+16)/2 is minimum when c=-3
so r=sqrt(7/2)
so g=f=-1/2
so equation of circle is x2+y2-x-y-3=0

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