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Find the equations of the tangents to the circle `x^2+y^2-6x+4y=12` which are parallel to the straight line `4x+3y+5=0`

A

`3x - 4y - 19 = 0 , 3x - 4y + 31 = 0`

B

`4x + 3y - 19 = 0 , 4x + 3y + 31 = 0`

C

`4x + 3y + 19 = 0 , 4x + 3y - 31 = 0`

D

`3x - 4y + 19 = 0 , 3x - 4y+ 31 = 0`

Text Solution

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The correct Answer is:
C
Let equation of tangent be 4x + 3y + k = 0
centre of the circle is (3 , -2)
Then , `sqrt(9 + 4 + 12) = |(4(3) + 3(-2) + k)/(sqrt(16 + 9))|`
`implies 6 + k = pm 25`
`implies k = 19` and `-31`
Hence , the tangent are 4x + 3y + 19 = 0 and 4x + 3y - 31 = 0 .
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