The number of circles belonging to the system of circles `2(x^(2)+y^(2))+lambda x-(1+lambda^(2))y-10=0` and orthogonal to `x^(2)+y^(2)+4x+6y+3=0`, is

Options

A. 2
B. 1
C. 0
D. none of these

Answer

A

Explanation

We have,
`x^(2)+y^(2)+(lambda)/(2)x-((1+lambda^(2))/(2)) y-5=0 and, x^(2)+y^(2)+4x+6y+3=0`
These two circles intersect orthogonally.
`:. 2 { (lambda)/(4)xx2-((1+lambda^(2))/(4))xx3}=-5+3`
`rArr lambda-(3)/(2)(lambda^(2)+1)=-2`
`rArr 2lambda-3lambda^(2)-3=-4`
`rArr 3lambda^(2)-2lambda-1=0rArr(3 lambda+1)(lambda-1)=0 rArr lambda=1, -(1)/(3)`
Each value of `lambda` gives a circle. Hence, there are two circles.