|=2 2h−3k+1=2√13 and √(h−5)2+(k−6)2=√13 From Eqs. (i) and (ii), we get 2h−3k+1=2√[(h−15)2+(k−6)2.] On squaring both sides, we get ‌4h2+9k2+1−12hk−6k +‌4h=4[h2+25−10h+k2+36−12k] ⇒5k2−12hk+44h+42k−243=0 ⇒12hk−5k2−44h−42k+243=0 Taking locus of P(h,k), we get 12xy−5y2−44x−42y+243=0