Find the equation of the circle passing through `(1,0) and (0,1)` and having the smallest possible radius.

Options

A. `x^(2)+y^(2)+x+y-2=0`
B. `x^(2)+y^(2)=x+y`
C. `x^(2)+y^(2)=1`
D. none of these

Answer

B

Explanation

Let the equation of the required circle be
`x^(2)+y^(2)+2gx+2fy+c=0 " " ...(i)`
This passes through A(1, 0) and B(0, 1). Therefore,
`1+2g+c=0` and, `1+2f+c=0`
`rArr g=-((c+1)/(2))` and, `f=-((c+1)/(2))`
Let r be the radius of circle (i). Then,
`r=sqrt(g^(2)+f^(2)-c)`

`rArr=sqrt(((c+1)/(2))^(2)+((c+1)/(2))^(2)-c)`
`rArr r=sqrt((c^(2)+1)/(2))rArr r^(2)=(1)/(2)(c^(2)+1)`
Clearly , r is minimum when c=0 and the minimum value of r is `(1)/(sqrt(2))`.
For c = 0, we have `g=-(1)/(2)` and `f=-(1)/(2)`
Substituting the values of g, f and c in (i), we get `x^(2)+y^(2)-x-y=0` as the equation of the required circle.