Let x_0, y_0 be fixed real numbers such that | vedclass

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(A) Let $P(x_0, y_0)$ be a fixed point outside the unit circle $x^2+y^2 \leq 1$.Let $Q(x, y)$ be any point on or inside the unit circle.The expression

Vedclass · Accepted Answer

$(\sqrt{x_0^2+y_0^2}-1)^2$

Vedclass · Answer

(A) Let $P(x_0, y_0)$ be a fixed point outside the unit circle $x^2+y^2 \leq 1$.Let $Q(x, y)$ be any point on or inside the unit circle.The expression $(x-x_0)^2+(y-y_0)^2$ represents the square of the distance $PQ^2$ between points $P$ and $Q$.Let $O$ be the origin $(0, 0)$. The distance $OP = \sqrt{x_0^2+y_0^2}$.Since $P$ is outside the circle,$OP > 1$.The distance $PQ$ is minimized when $Q$ lies on the line segment $OP$.In this case,the distance $PQ = OP - OQ$.Since the minimum distance $PQ$ occurs when $Q$ is on the boundary of the circle,we have $OQ = 1$.Thus,the minimum distance $PQ = \sqrt{x_0^2+y_0^2} - 1$.The minimum value of $PQ^2$ is $(\sqrt{x_0^2+y_0^2}-1)^2$.

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

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