The straight line `x/a+y/b=1` cuts the coordinate axes at `A` and `B` . Find the equation of the circle passing through `O(0,0),Aa n dBdot`

Options

A. `x^(2)+y^(2)-ax-by=0`
B. `x^(2)+y^(2)-2ax-2by=0`
C. `x^(2)+y^(2)+ax+by=0`
D. `x^(2)+y^(2)=a^(2)+b^(2)`

Answer

A

Explanation

The straight line `(x)/(a)+(y)/(b)=1` cuts the coordinate axes at A(a, 0) and B(0, b).
Let `x^(2)+y^(2)+2gx+2fy+c=0 " " ...(i)`
be the circle passing through O, A and B. Then,
`0+c=0 " " (ii)`
`a^(2)+2ga+c=0 " " ...(iii)`
`b^(2)+2fb+c=0 " " (iv)`
Solving (ii), (iii) and (iv), we obtain
`g=-(a)/(2), f=-(b)/(2)` and `c=0`
Substituting these values in (i), we obtain the equation of the required circle as `x^(2)+y^(2)-ax-by=0`