TS EAMCET 9 May 2024 Shift 1 Solved Paper

We have,
‌S1:x2+y2−4x+6y+4=0
‌C1:(2−3),r1=3
‌S2:x2+y2+2x−2y−2=0
‌C2:(−1,1),r2=2
Here, C1C2=r1+r2
So, the point P divides C1C2 in the ratio of 3:2
P(‌

1

5

,‌

−3

5

)
Slope of line passing through C1 and C2 is ‌

1+3

−1−2

=‌

4

−3

⇒P:(‌

1

5

,−‌

3

5

) and here only one transverse common tangent which is perpendicular to C1C2
‌ ⇒ slope of tangent ‌=−‌

1

−‌ 4 3

=‌

3

4

∵ Tangent is also passes through (‌

1

5

,−‌

3

5

)
So, equation of tangent is
y+‌

3

5

=‌

3

4

(x−‌

1

5

)⇒3x−4y−3=0
On comparing with
ax+by+c=0,‌

a

c

=−1