Let co-ordinates of P are (x,y). Since, P is equidistant from A,B,C, then ‌PA2=PB2 . . . (i) and ‌PB2=PC2 . . . (ii) ‌‌ From Eq. (i), ‌ ‌(x−1)2+(y−3)2=(x+3)2+(y−5)2 ‌⇒‌‌x2+1−2x+y2+9−6y ‌=x2+9+6x+y2+25−10y ‌⇒‌‌8x−4y+24=0 ‌⇒‌‌2x−y+6=0 . . . (iii) ‌‌ From Eq. (ii), ‌ ‌(x+3)2+(y−5)2=(x−5)2+(y+1)2 ‌⇒‌‌x2+9+6x+y2+25−10y ‌=x2+25−10x+y2+1+2y ‌⇒‌‌16x−12y+8=0 ‌⇒‌‌4x−3y+2=0 . . . (iv) On solving Eqs. (iii) and (iv), we get x=−8,y=−10 Now, PA2=(−8−1)2+(−10−3)2 ‌=81+169=250 PA‌=√250=5√10