JEE Advanced2010MathematicsCircleActual
Paragraph: The circle x^2+y^2-8 x=0 and hyperbola x^2 9 - y^2 4 =1 intersect at the points A and B . Question: Equation of the circle with A B as its diameter is
Options
- Ax^2+y^2-12 x+24=0
- Bx^2+y^2+12 x+24=0
- Cx^2+y^2+24 x-12=0
- Dx^2+y^2-24 x-12=0
Correct answer
A. x^2+y^2-12 x+24=0
Step-by-step solution
The equation of the hyperbola is x^2 9 - y^2 4 =1 and that of circle is x^2+y^2-8 x=0 For their points of intersection x^2 9 + x^2-8 x 4 =1 array ll & 4 x^2+9 x^2-72 x=36 & 13 x^2-72 x-36=0 array aligned & 13 x^2-78 x+6 x-36=0 & 13 x(x-6)+6(x-6)=0 & x=6, x=- 13 6 & x=- 13 6 not acceptable & aligned Now, for x=6, y= 2 3 Required equation is aligned & (x-6)^2+(y+2 3 )(y-2 3 )=0 & x^2-12 x+y^2+24=0 & x^2+y^2-12 x+24=0 aligned