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If a line, y = mx + c is a tangent to th...

If a line, y = mx + c is a tangent to the circle, `(x-1)^2 + y^2 =1` and it is perpendicular to a line `L_1` , where `L_1` is the tangent to the circle `x^2 + y^2 = 8` at the point (2, 2), then :

A

`c^(2) + 2c -1 = 0`

B

`c^(2) - 2c+1= 0`

C

`c^(2) -2c-1 = 0`

D

`c^(2) + 2c + 2=0`

Text Solution

Verified by Experts

The correct Answer is:
A
Slope of tangent to `x^2 + y^2 = 8` at P (2,2)
`2x + 2y y. = 0 rArr m_(T|P)=-1`
`y = mx + c` is tangent to `(x-1)^2 + y^2 =1`
`y=x+c` is tangent to `(x-1)^2 + y^2 =1`
`|(c+1)/(sqrt2)|=1 rArr c^(2) + 2c -1 =0`
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