Question

The members of a family of circles are given by the equation `2(x^2 + y^2) + 2x-(1+lambda^2)y-10=10.` The number of circles belonging to the family that are cut orthogonally by the fixed circle `x^2+y^2+4x +6y + 3 = 0` is
A. 2
B. 1
C. 0
D. none of these

Answer 1

Correct Answer - A
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.