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Shortest Distance between a Point and a Circle

What is the distance between a circle $C$ with equation $x^{2} + y^{2} = r^{2}$ which is centered at the origin and a point $P (x_{1}, y_{1})$ ?

The ray $\vec{O P}$ , starting at the origin $O$ and passing through the point $P$ , intersects the circle at the point closest to $P$ . So, the distance between the circle and the point will be the difference of the distance of the point from the origin and the radius of the circle.

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Using the Distance Formula , the shortest distance between the point and the circle is $| \sqrt{{(x_{1})}^{2} + {(y_{1})}^{2}} - r |$ .

Note that the formula works whether $P$ is inside or outside the circle.

If the circle is not centered at the origin but has a center say $(h, k)$ and a radius $r$ , the shortest distance between the point $P (x_{1}, y_{1})$ and the circle is $| \sqrt{{(x_{1} - h)}^{2} + {(y_{1} - k)}^{2}} - r |$ .

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Example 1:

What is the shortest distance between the circle $x^{2} + y^{2} = 9$ and the point $A (3, 4)$ ?

The circle is centered at the origin and has a radius $3$ .

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So, the shortest distance $D$ between the point and the circle is given by

$\begin{array}{l} D = | \sqrt{{(3)}^{2} + {(4)}^{2}} - 3 | \\ = | \sqrt{25} - 3 | \\ = | 5 - 3 | \\ = 2 \end{array}$

That is, the shortest distance between them is $2$ units.

Example 2:

What is the shortest distance between the circle $x^{2} + y^{2} = 36$ and the point $Q (- 2, 2)$ ?

The circle is centered at the origin and has a radius $6$ .

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So, the shortest distance $D$ between the point and the circle is given by

$\begin{array}{l} D = | \sqrt{{(- 2)}^{2} + {(2)}^{2}} - 6 | \\ = | \sqrt{8} - 6 | \\ = 6 - 2 \sqrt{2} \\ \approx 3.17 \end{array}$

That is, the shortest distance between them is about $3.17$ units.

Example 3:

What is the shortest distance between the circle ${(x + 3)}^{2} + {(y - 3)}^{2} = 5^{2}$ and the point $Z (- 2, 0)$ ?

Compare the given equation with the standard form of equation of the circle,

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ where $(h, k)$ is the center and $r$ is the radius.

The given circle has its center at

(- 3, 3)

and has a radius of

5

units.

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Then, the shortest distance $D$ between the point and the circle is given by

$\begin{array}{l} D = | 5 - \sqrt{{(- 3 - (- 2))}^{2} + {(3 - 0)}^{2}} | \\ = | 5 - \sqrt{1 + 9} | \\ = | 5 - \sqrt{10} | \\ \approx 1.84 \end{array}$

That is, the shortest distance between them is about $1.84$ units.

Example 4:

What is the shortest distance between the circle $x^{2} + y^{2} - 8 x + 10 y - 8 = 0$ and the point $P (- 4, - 11)$ ?

Rewrite the equation of the circle in the form ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ where $(h, k)$ is the center and $r$ is the radius.

$\begin{array}{l} x^{2} + y^{2} - 8 x + 10 y - 8 = 0 \\ x^{2} - 8 x + 16 + y^{2} + 10 y + 25 = 8 + 16 + 25 \\ {(x - 4)}^{2} + {(y + 5)}^{2} = 49 \\ {(x - 4)}^{2} + {(y + 5)}^{2} = 7^{2} \end{array}$

So, the circle has its center at

(4, - 5)

and has a radius of

7

units.

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Then, the shortest distance $D$ between the point and the circle is given by

$\begin{matrix} D = | \sqrt{{(- 4 - 4)}^{2} + {(- 11 - (- 5))}^{2}} - 7 | \\ = | \sqrt{64 + 36} - 7 | \\ = \sqrt{100} - 7 \\ = 3 \end{matrix}$

That is, the shortest distance between them is $3$ units.