Question

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 $x^2+y^2\le 1$, then the minimum value of $(x–x_0{)}^2+(y–y_0{)}^2$ is

• 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. ${(|x_0|+|y_0|)}^2-1$

Answer 1

The correct option is A ${(\sqrt{x_0^2+y_0^2}-1)}^2$

Given :
$x_0^2+y_0^2>1$ and $x^2+y^2\le 1$

$x_0,y_0$ are points outside the circle
$x,y$ are points on the circle or inside the circle.

$(x–x_0{)}^2+(y–y_0{)}^2=PQ$,
$\therefore$ it will be minimum when P, Q , C all are collinear and Q lies on the circle.
Min. value will be
$=(x–x_0{)}^2+(y–y_0{)}^2=(PC-QC{)}^2={(\sqrt{x_0^2+y_0^2}-1)}^2$