The equation of the tangent from the point (0, 1) to the circle `x^2+y^2-2x-6y+6=0`

Options

A. ` 3 ( x ^ 2 - y^ 2 ) + 4x y - 4x - 6y + 3 = 0 `
B. ` 3y ^ 2 + 4x y - 4x -6y + 3 = 0 `
C. ` 3 x ^2 + 4xy - 4x - 6y + 3 = 0 `
D. ` 3 ( x ^2 + y ^ 2 ) + 4xy - 4x - 6y + 3 = 0 `

Answer

B

Explanation

Let ` S -= x ^ 2 + y ^ 2 -2x -6y + 6 = 0 `
and ` P ( x _ 1 , y _ 1 ) = ( 0, 1 ) `
` S _ 1 = x_ 1 ^ 2 + y _ 1 ^ 2 - 2 x _ 1 - 6y _ 1 + 6 = 0`
` = ( 0 ) ^ 2 + (1 ) ^ 2 - 2 ( 0) -6 ( 1 ) + 6 `
` = 1 -6 + 6 = 1 `
` T = x * x _ 1 + y * y _ 1 - ( x - x _ 1 ) - 3 ( y- y _ 1 ) + 6 `
` = x * ( 0 ) + y ( 1 ) - ( x + 0 ) - 3 ( y + 1 ) + 6 `
` =0 + y- x - 3y - 3 + 6 `
` = - x - 2y + 3 `
` rArr T ^ 2 = ( - x - 2y + 3 ) ^ 2 `
` = ( - x- 2 y ) ^ 2 + 9 + 6 ( - x - 2 y ) `
` = x^ 2 + 4y ^ 2 + 4xy + 9 - 6 x - 12 y `
` therefore ` Equation of the pair of tangents
` S * S _ 1 = T ^ 2 `
` ( x ^ 2 + y^ 2 - 2x - 6y + 6) ( 1 ) `
` = x^ 2 + 4y ^ 2 + 4xy - 6x - 12 y + 9 `
` rArr 3y ^ 2 + 4xy -4x - 6y + 3 = 0 `