Given, AB+BC+AC=20 ⇒AB+AC=9 ⇒√(h+5)2+k2+√(h−6)2+k2=9 ⇒√(h+5)2+k2=9−√(h−6)2+k2 On squaring both sides, h2+10h+25+k2=81+(h2−12h+36+k2) ⇒2×9×√h2−12h+36+k2 −2⋅9√(h−6)2+k2 =−22h+92=2(−11h+46) ⇒9√h2−12h+36+k2=−11h+46 Squaring on both sides, 81(h2−12h+36+k2)=121h2−1012h+2116 ⇒40h2−40h−81k2−800=0 ∴ Locus of the third vertex is 40x2−81y2−40x−800=0