Which point on the first circle is closest to the second?

Which point on the first circle is closest to the second?

Question

The point on the circle $(x-5)^2+(y-4)^2=4$ which is closest to the circle $(x-1)^2+(y-1)^2=1$ is

1. $(3.4,2.8)$,

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1. $(3,4)$,

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1. $(5,2)$,

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1. $(3.8,2.4)$.

Solution

Here is a diagram of the circles:

Approach 1

The right-angle triangle with the circle centres at two of its vertices has sides of length $3$ and $4$. Its hypotenuse is therefore of length $5$.

Using similar triangles, we have $\dfrac{3}{5}=\dfrac{a}{4}=\dfrac{b}{3}$. Thus $a = 2.4, b = 1.8$, and the point on the first circle closest to the second is $(3.4,2.8)$. The answer is (a).

Approach 2

The vector joining $(5,4)$ and $(1,1)$, the centres of the circles, is $\begin{pmatrix}-4\\-3\end{pmatrix}$. The length of this vector is $\sqrt{(-4)^2+(-3)^2}=5$.

The first circle has radius $2$, and so the point on this closest to the second is $\dfrac{2}{5}$ths of the way along this vector. Hence we can calculate the point: $\begin{pmatrix}5\\4\end{pmatrix} + \frac{2}{5}\begin{pmatrix}-4\\-3\end{pmatrix} = \begin{pmatrix}5\\4\end{pmatrix} + \begin{pmatrix}-1.6\\-1.2\end{pmatrix} = \begin{pmatrix}3.4\\2.8\end{pmatrix}.$

Therefore the answer is (a).