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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 -2 c +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+2yy'=0 " "rArr" "mr = -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`
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