AP EAPCET 24-Aug-2021 Shift 1 Paper

C1:x2+y2+x+y−4=0
C2:3x2+3y2−x−y−λ=0
⇒x2+y2−

x

3

−

y

3

−

λ

3

=0
∴ Length of tangent drawn by an external point is √S1.
∴ Length of tangent drawn from (1,2) to C1 and C2 are L1 and L2 respectively.
L1=√S1=√12+22+1+2−4
=√1+4+1+2−4=√4
L1=2
∴L2=√S12=√12+22−

1

3

−

2

3

−

λ

3

=√4−

λ

3

Given that,

L1

L2

=

3

4

=

2

√4− λ 3

=

3

4

⇒3√4−

λ

3

=8
⇒9(4−

λ

3

)=64
⇒4−

λ

3

=

64

9

⇒4−

64

9

=

λ

3

⇒

36−64

9

=

λ

3

λ=−

28

3