TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the figure shown below
The circles are,
x2+y2+4x+4y−10=0
Its centre is (-2,-2)
And,
x2+y2−6x−6y+10=0
Its centre is (3,3)
Now P is the point of contact of circle
P‌‌=(‌

9√2−4√2

5√2

,‌

9√2−4√2

5√2

)
‌‌=(1,1)
And Q is external similitude
Q=(‌

9√2+4√2

5√2

,‌

9√2+4√2

5√2

)
=(13,13)
Therefore, the equation of circle with PQ as diameter is
(x−1)(x−13)+(y−1)(y−13)‌‌=0
x2+y2−14x−14y+26‌‌=0