TG EAMCET 3 May 2025 Shift 2 Paper

We have two circles given by:
‌S1:x2+y2−4x−6y−12=0
‌S2:x2+y2+6x+18y+26=0
We need to find a third circle that:
Touches both S1 and S2 at their point of contact
Passes through the point (1,−1)
Any circle that touches both S1 and S2 at their point of contact can be written as:
(x2+y2−4x−6y−12)+λ(x2+y2+6x+18y+26)=0
Since the circle also passes through the point (1,−1), substitute x=1,y=−1 into the equation: (1)2+(−1)2−4(1)−6(−1)−12+λ[(1)2+(−1)2+6(1)+18(−1)+26]=0
Calculate each term step-by-step:
First bracket: 1+1−4−(−6)−12=2−4+6−12=−8 (Notice −6×−1=+6 )
Second bracket: 1+1+6−18+26=2+6−18+26=8−18+26=−10+26=16
So the equation becomes:
−8+λ(16)=0
Solve for λ :
λ=‌

8

16

=‌

1

2

Now plug λ=‌

1

2

into our general circle equation:
‌(x2+y2−4x−6y−12)+‌

1

2

(x2+y2+6x+18y+26)=0
‌x2+y2−4x−6y−12+‌

1

2

x2+‌

1

2

y2+3x+9y+13=0
‌(1+‌

1

2

)x2+(1+‌

1

2

)y2+(−4+3)x+(−6+9)y+(−12+13)=0

‌

3

2

x2+‌

3

2

y2−x+3y+1=0
To make it simpler, multiply both sides by 2 :
3x2+3y2−2x+6y+2=0
Divide all terms by 3 to get standard form:
x2+y2−‌

2

3

x+2y+‌

2

3

=0
The center of a circle x2+y2+2gx+2fy+c=0 is (−g,−f).
From the equation:
‌2g=−‌

2

3

‌ so ‌g=−‌

1

3

‌2f=2‌ so ‌f=1
Therefore, the center is (‌

1

3

,−1).