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