The members of a family of circles are given by the equation 2 (x2+ y2) + λx -(1 + λ2) - Maths - Conic Sections

Question

The members of a family of circles are given by the
equation 2 (x2+ y2) + λx -(1 + λ2) y -10 = 0 The
number of circles belonging to the family that are cut orthogonally
by the fixed circle x2 + y2 + 4x + 6y + 3 = 0 is
(A) 2
(B) 1
(C) 0
(D) none of these

Aarushi Mishra · Accepted Answer

C1: 2x2+2y2+λx-(1+λ2)y-10=0:x2+y2+λ2x-(1+λ22)y-5=0g1=λ4f1=-1+λ24c1= -5C2:x2+y2+4x+6y+3=0g2=2f2=3c2=3Condition for orthogonality of two circles2g1g2 + 2f1f2 = c1+c22×λ4×2 - 2×1+λ24×3 = -5 +3λ-3+3λ22 =-22λ-3-3λ2=-43λ2-2λ-1=0Discriminant D=b2 - 4acD=22+4×3×1>0Hence it has two real roots
Ans A

The members of a family of circles are given by the equation 2 (x2+ y2) + λx -(1 + λ2) y -10 = 0. The number of circles belonging to the family that are cut orthogonally by the fixed circle x2 + y2 + 4x + 6y + 3 = 0 is (A) 2 (B) 1 (C) 0 (D) none of these

The members of a family of circles are given by the equation 2 (x2+ y2) + λx -(1 + λ2) y -10 = 0. The number of circles belonging to the family that are cut orthogonally by the fixed circle x2 + y2 + 4x + 6y + 3 = 0 is

(A) 2

(B) 1

(C) 0

(D) none of these