TS EAMCET 9 May 2024 Shift 1 Solved Paper

We have,
C1:x2+y2−2x−8y+8=0
Centre (1,4),r1=3
C2:x2+y2−8x+15=0
Centre (4,0),r2=1
As ‌

C1P

C2P

=‌

C1A

C2C

=‌

3

1

⇒Px=‌

12−1

2

=‌

11

2

‌‌ (external division)
Py=‌

0−4

2

=−2
Let slope of tangent AP be m.
Then, equation of AP is :
y+2=m(x−‌

11

2

)
2mx−2y−(4+11m)=0
∵ Perpendicular distance of C2 from AP is 1 .
‌∴|‌

2m×4−2×0−(4+11m)

√(2m)2+(−2)2

|=1
‌(3m+4)2=4m2+4
‌5m2+24m+12=0
So, m1+m2=−‌

24

5

where m1 and m2 are slope of AP and BP.