Q32 / 56 · KVPY 2018 (SA) · Mathematics

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 $x^{2} + y^{2} \leq 1,$ then the minimum value of ${(x - x_{0})}^{2} + {(y - y_{0})}^{2}$ is

Correct answer
A
${(\sqrt{x_{0}^{2} + y_{0}^{2}} - 1)}^{2}$
B
$x_{0}^{2} + y_{0}^{2} - 1$
C
${(|x_{0}| + |y_{0}| - 1)}^{2}$
D
${(|x_{0}| + |y_{0}|)}^{2} - 1$

Solution

$Let P (x_{0}, y_{0})$

Given $x^{2} + y^{2} \leq 1$ 1$

Let any arbitrary point $8 (x, y)$ .

$P Q^{2} = {(x - x_{0})}^{2} + {(y - y_{0})}^{2}$ $P Q^{2} = (O P - O Q)^{2}$
$P Q^{2} = (O P - O Q)^{2}$ $P Q^{2} = {(\sqrt{x_{0}^{2} + x_{0}^{2}} - 1)}^{2} [∵ 0 Q = 1]$

$∴$ Mininimum value of $P Q^{2}$ is

${(\sqrt{x_{0}^{2} + y_{0}^{2}} - 1)}^{2}$

Hence, the correct answer is option: 'A'.

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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 $x^{2} + y^{2} \leq 1,$ then the minimum value of ${(x - x_{0})}^{2} + {(y - y_{0})}^{2}$ is