AP EAMCET20228 Jul 2022Morning ShiftMathematicsCircleActual
The straight line x + y = p cuts the circle x^2+y^2-a^2=0 at A and B . Then the equation of circle having AB as diameter is
Options
- Ax^2+y^2-a^2+p(x a+y a-p)=0
- Bx^2+y^2-a^2-p(x +y +p)=0
- Cx^2+y^2-a^2+2 p(x a+y a-p)=0
- Dx^2+y^2-a^2-2 p(x +y -p)=0
Correct answer
D. x^2+y^2-a^2-2 p(x +y -p)=0
Step-by-step solution
Since Centre of the circle having diameter as AB. Now foot of perpendicular from (0,0) to the line x + y = p is aligned & x -0 = y -0 = (- P ) 1 & x = p , y = P aligned how equation of circle having centre ( P , P ) and radius (r)= a^2-p^2 is ( x - p )^2+( y - p )^2= ( a ^2- p )^2 x ^2+ y ^2- a ^2-2 p ( x + y - p )=0