$\text{ Let }AB\text{ be the chord of contact of the point }(5,-5)\text{ with respect to the circle }{x}^2+y^2=5\text{ . }$

$\text{ Then the locus of the orthocenter of }\mathrm{\Delta }PAB,\text{ where }P\text{ is any point on the circle, is }$

detailed solution

Correct option is A

Equation of chord of contact  AB will be S1=0 i.e ,5x−5y=5⇒x−y=1 Let Ax1,x1−1,Bx2,x2−1 be the points  Let P be (5cos⁡θ,5sin⁡θ)since A,B, P lies on the circle ,centre O0,0 is the circumcentre of the ∆PAB and centroid is 5cos⁡θ+x1+x23+5sin⁡θ+x1−1+x2−13    ⇒ Ortho centre will be 5cos⁡θ+x1+x2,5sin⁡θ+x1+x2−2=(h,k)⇒5cos⁡θ=h−x1+x2,5sin⁡θ=k+2-x1+x2since (5cos⁡θ)2+(5sin⁡θ)2=5   if circumcentre is the origin then Orhocentre =3 G =x1+x2+x3,y1 +y2+y3⇒h−x1+x22+k+2−x1+x22=5…… (1)x1,x1−1,x2,x2−1  lies on circle x2+(x−1)2=5x2−x−2=0⇒x1+x2=1∴ Locus will be equation (1):(h−1)2+(k+1)2=5⇒(x−1)2+(y+1)2=5