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The equation of a circle of radius 1 tou...

The equation of a circle of radius `1` touching the circles `x^2+y^2-2|x|=0` is (a) `x^2+y^2+2sqrt(2)x+1=0` (b) `x^2+y^2-2sqrt(3)y+2=0` (c) `x^2+y^2+2sqrt(3)y+2=0` (d) `x^2+y^2-2sqrt(2)+1=0`

A

`y=4sqrt(5)x+20`

B

`y=4sqrt(3)x-12`

C

y=0

D

`y=-4sqrt(5)x-20`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

1,2,3

`y=x|x|={{:(x^(2)","xge0),(-x^(2)","xlt0):}`
y=mx+c
If it touches `y=x^(2),xge0` then `mx+c=x^(2)` has equal roots, i.e.,
`m^(2)+4c=0orc=(-m^(2))/(4)` (2)
Also, the line touches the circle. Therefore,
`(|0-2+c|)/(sqrt(m^(2)+1))=2` (3)
`or4-4c+c^(2)=4(1-4c)`
`c^(2)+12c=0`
`i.e.," "c=0or-12`
For c=0, m=0
Therefore, the equation of tangent is `y=sqrt(48)x-12`.
If line (1) touches `y=-x^(2),xlt0`, then the equation `mx+c=-x^(2)` has equal roots. Therefore,
D=0
`orm^(2)-4c=0` (4)
From (3) and (4), we get
`c^(2)20c=0`
`:.c=0orc=20`
For `c=20, m=sqrt(80)`
Therefore, the equation of tangent is `y=sqrt(80)x+20`
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