TS EAMCET 5-Aug-2021 Shift 2 Question Paper

C1:(x−1)2+(y+9)2=4
C2:(x+4)2+(y−3)2=225
Let's at first solve the equations
10x−24y−278=0
5x−12y−139=0
Let's also check the perpendicular distance from centre of both the circles. From C1:‌

|5⋅1−12⋅(−9)−139|

13

=‌

26

13

=2
From C2:‌

|(5⋅−4)−(−12⋅3)−225|

13

=‌

195

13

=15
As the perpendicular distance from both the centre of the circles is same as the length of radius. So, this line is only the common tangent.
5x−12y−139=0
5(‌

−2

5

)−12(‌

−47

4

)−139=0
Option (d) satisfies the equation, so we can say that (d) is the correct option.