The members of a family of circles are given by the equation 2(x²+y²)+ λx-(1+λ²)y-10=0.The no. 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

Hi Menka,

Use the condition for orthogonality.

2gg' + 2ff' = c+c' --------Use this condition only after writing the two cirlce eqns in standard form (Co-ef of sq terms = 1).

Using this condition, you'll get a quad eqn in λ.

The no of roots, which is the number of circles, will be 0,1, or 2 depending on the D of the QE accordinlgy as <0, =0, or >0.

Best Regards,

Ashwin (IIT Madras).

We Know tht the conditions for circles to intersect orthogonally is:

2(g1g2+f1f2)=c1+c2

now applying this we get.

2(-2*-λ+-3*(1+λ²)/2)=3-10

solving ..

λ=2,-2/3
hence their r two  circles ..