The equation of the smallest circle passing from points (1, 1) and (2, 2) and always in the first quadrant is

Options

A. `x^(2)+y^(2)-4x-2y+4=0`
B. `x^(2)+y^(2)+2x+4y+4=0`
C. `x^(2)+y^(2)-3x-3y+4=0`
D. `x^(2)+y^(2)-5x-y+4=0`

Answer

C

Explanation

Let the equation of the required circle be
`x^(2)+y^(2)+2gx+2fy+c=0 " " ...(i)`
It passes through (1, 1) and (2,2 ).
`:. 2+2g+2f+c=0 " " ...(ii)`
`8+4g+4f+c=0 " " ...(iii)`
From (ii) and (iii), we get
g+f = -3 and c=4.
Let r be the radius of the circle . Then,
`r^(2)=g^(2)+f^(2)-c`
`rArr r^(2)=g^(2)+f^(2)-4`
`rArr r^(2)=g^(2)+(3+g)^(2)-4 " "[:' g+f=-3]`
`rArrr^(2)=2g^(2)+6g+5`
`rArr r^(2)=2{g^(2)+3g+(5)/(2)}rArr r^(2)=2 {(g+(3)/(2))^(2)+(1)/(4)}`
Clearly, r is least when g `+(3)/(2)`=0 i.e. `g=-(3)/(2)`
Putting `g=-(3)/(2)` in g+f = -3, we get `f=-(3)/(2)`
Substituting the values of g, f and c in (i), we get
`x^(2)+y^(2)-3x-3y+4=0`