The equation of common tangent of `x^2 + y^2=2 and y^2 = 8x` is

Options

A. y=x+1
B. y=x+2
C. y=x-2
D. y=-x+2

Answer

B

Explanation

Given `y^(2)=8x`
`:. 4a=8 rArr a=2`
Any tangent of parabola is, `y=mx+(a)/(m)`
`rArr mx-y+(2)/(m)=0`
If is a triangle to the circle `c^(2)j+y^(2)=2`. Then perpendicular from centre (0,0) is equal to radius `(sqrt(2))/(2)`
`:. (m)/(sqrt(m^(2)+1))=sqrt(2)`
`rArr (4)/(m^(2))=2(m^(2)+1)`
`rArr m^(2)+m^(2)-2=0`
`(m^(2)+2)(m^(2)-1)=0`
`m+-1`
Hence, the common tangents are `y=+-(x+2)`