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

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

Grade:12th Pass

2 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

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).

Jayesh Muley
20 Points
9 years ago

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 ..


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