Highly selective Backlog Qs for JEE MainMathematicsCircle
Let x 0 , y 0 be fixed real numbers such that x 0 2 + y 0 2 > 1 . If x , y are arbitrary real numbers such that x 2 + y 2 ≤ 1 , then the minimum value of x - x 0 2 + y - y 0 2 is
Options
- Ax 0 2 + y 0 2 - 1 2
- Bx 0 2 + y 0 2 - 1
- Cx 0 + y 0 - 1 2
- Dx 0 + y 0 2 - 1
Correct answer
A. x 0 2 + y 0 2 - 1 2
Step-by-step solution
Let P x 0 , y 0 Given x 2 + y 2 ≤ 1 1 Let any arbitrary point 8 ( x , y ) . P Q 2 = x - x 0 2 + y - y 0 2 P Q 2 = ( O P - O Q ) 2 P Q 2 = ( O P - O Q ) 2 P Q 2 = x 0 2 + x 0 2 - 1 2 [ ∵ 0 Q = 1 ] ∴ Mininimum value of P Q 2 is x 0 2 + y 0 2 - 1 2