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