Answer
$P_{1}(3,6)$ is the closest point and $P_{2}(-3,-6)$ the furthest point.
Work Step by Step
The distance from $P(x, y)$ to $Q(1,2)$ is given by
\[
d(x, y)=\sqrt{(x-1)^{2}+(y-2)^{2}}=d
\]
Using Lagrange multipliers under the constraint $45=x^{2}+y^{2},\mathbf{P}$ is on the circle
\[
\begin{aligned}
\nabla d=\lambda \nabla g & \Rightarrow \frac{2}{2 d}(x-1) \mathrm{i}+\frac{2}{2 d}(y-2) \mathrm{j}=\lambda(2 x \mathrm{i}+2 y \mathrm{j}) \\
& \Rightarrow \lambda=\frac{x-1}{2 d x}=\frac{y-2}{2 d y} \\
& \Rightarrow \frac{x-1}{x}=\frac{y-2}{y} \Rightarrow y=2 x
\end{aligned}
\]
Using this in the given constraint:
\[
\begin{array}{c}
x^{2}+y^{2}=45 \Rightarrow x^{2}+4 x^{2}=45 \Rightarrow x^{2}=9 \Rightarrow x=\pm 3 \\
y=2 x=\pm 6
\end{array}
\]
Hence, $P_{1}(3,6)$ and $P(-3,-6)$ are the absolute extrema. We note that
\[
d(3,6)=20 \quad \text { and } \quad d(-3,-6)=80
\]
Therefore $P_{1}(3,6)$ is the closest point and $P_{2}(-3,-6)$ the furthest point.
