JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 17)
The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d2 is equal to _____________.
Answer
16
Explanation
Let point P(x, y)
A(0, 0), B(1, 0), C(0, 1), D(1, 1)
(PA)2 + (PB)2 + (PC)2 + (PD)2 = 18
$${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$$ = 18
$$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$$
$$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$$
$$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $$
$$ \Rightarrow {d^2} = 16$$
A(0, 0), B(1, 0), C(0, 1), D(1, 1)
(PA)2 + (PB)2 + (PC)2 + (PD)2 = 18
$${x^2} + {y^2} + {x^2} + {(y - 1)^2} + {(x - 1)^2} + {y^2} + {(x - 1)^2} + {(y - 1)^2}$$ = 18
$$ \Rightarrow 4({x^2} + {y^2}) - 4y - 4x = 14$$
$$ \Rightarrow {x^2} + {y^2} - x - y - {7 \over 2} = 0$$
$$d = 2\sqrt {{1 \over 4} + {1 \over 4} + {7 \over 2}} $$
$$ \Rightarrow {d^2} = 16$$
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