AP EAPCET 24-Aug-2021 Shift 1 Paper

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

x

3

−

y

3

−

k

3

=0
Length of tangents drawn from external point to the circle is √S1.
Now, according to the question,
Let length of tangents drawn from point (1,2) to circles C1 and C2 are L1 and L2 respectively.
So, L1=√S1=√12+22+1+2−4
=√1+4+3−4=√4
L2=√S1′=√12+22−

1

3

−

2

3

−

k

3

=√1+4−1−

k

3

=√4−

k

3

Now, given

L1

L2

=

4

3

⇒

√4

√4− k 3

=

4

3

⇒

4

4− k 3

=

16

9

[squaring]
⇒9=4(4−

k

3

)
⇒

9

4

=4−

k

3

k

3

=4−

9

4

k

3

=

7

4

⇒k=

21

4