The equation of the line that touches the curves `y=x|x|` and `x^2+(y^2-2)^2=4` , where `x!=0,` is `y=4sqrt(5)x+20` (b) `y=4sqrt(3)x-12` `y=0` (d) `y=-4sqrt(5)x-20`

Options

A. `y=4sqrt(5)x+20`
B. `y=4sqrt(3)x-12`
C. y=0
D. `y=-4sqrt(5)x-20`

Answer

A, B, C

Explanation

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`