As two sides are along co−ordinate tomahawks. Henceforth triangle OAB will
be correct calculated.
Leave locus alone circumcentre be (h,k) and condition of hypotenuse be xp+yq=1.
As triangle is correct calculated henceforth circumcentre will be at the mid purpose of the hypotenuse.
⇒(h,k)=(p+0/2,0+q/2)
⇒p=2h and q=2k.
⇒x/p+y/q=1 moves toward becoming x/2h+y/2k=1
x/h+y/k=2 (i)
Given that hypotenuse goes through ax+by+c=0 ii
lx+my+n=0 iii
(equation(ii)×l −equation(iii)×a, we get,alx+bly+cl−alx−amy−an=0
⇒y(bl−am)=an−cl⇒y=an−cl/bl−am
equation(ii)×m −equation(iii)×b, we get,amx+bmy+cm−blx−bmy−bn=0
x(am−bl)=bn−cm
⇒x=bn−cmam−bl
Putting the estimation of (x,y) in equation(i), we get,1h(bn−cmam−bl)+1k(an−clbl−am)=2
⇒k(bn−cm)(bl−am)+h(an−cl)(am−bl)hk(am−bl)(bl−am)=2
⇒k(bn−cm)(bl−am)+h(an−cl)(am−bl)=2hk(am−bl)(bl−am)
Replace h→x and k→y to get the condition of locus of circumcenter.
⇒y(bn−cm)(bl−am)+x(an−cl)(am−bl)=2xy(am−bl)(bl−am)
