The equation of the circle which passes the origin and cuts orthogonally each of the circles ` x^(2) + y^(2) - 6x + 8 = 0 " and " x^(2) + y^(2) - 2x - 2y = 7 ` is

Options

A. `3x^(2) + 3y^(2) = 8x - 13y = 0`
B. `3x^(2) + 3y^(2) - 8x + 29y = 0`
C. `3x^(2) + 3y^(2) + 8x + 29y = 0`
D. `3x^(2) + 3y^(2) - 8x - 29 y = 0`

Answer

B

Explanation

Given circles are, `S -= x^(2) + y^(2) - 6x + 8 = 0`
and `S^(1) -= x^(2) + y^(2) - 2x - 2y - 7 = 0`
Let the required circle is `x^(2) + y^(2) + 2gx + 2fy + c = 0`
Equation (1) cuts S = 0 orthogonally then
2 (-3) g + 0 c + g
since equation (1) passes through origin c = 0
`implies - 6g = 8`
`implies g = - (4)/(3)`
`implies` Equation (1) cuts `S^(1) = 0` orthogonally then,
2 (-1)(g) + 2 (-1)(F) - 0 - 7
`- 2 ((-4)/(3)) - 2f = - 7`
`implies -2f = - 7 - (8)/(3) = (-29)/(3)`
`implies f = (29)/(6)`
Equation (1) `implies x^(2) + y^(2) - (8)/(3) + (29)/(3) y = 0`
`3x^(2) + 3y^(2) - 8x + 29y = 0`