The correct option is C x2+y2–12x+24=0
The equation of the hyperbola is
x29−y24=1
and that of circle is
x2+y2–8x=0
For their points of intersection
x29+x2−8x4=1
⇒4x2+9x2–72x=36
⇒13x2–72x–36=0
⇒13x2–78x+6x–36=0
⇒13x(x–6)+6(x–6)=0
⇒x=6,x=−136
x=−136 not acceptable
Now, for x=6,y=±2√3
Required equation is
(x−6)2+(y+2√3)(y−2√3)=0
⇒x2–12x+y2+24=0
⇒x2+y2–12x+24=0