# 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 $P\left({x}_{1},{y}_{1}\right)$ ?

The ray $\stackrel{\to}{OP}$ , 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.

Using the Distance Formula , the shortest distance between the point and the circle is $\left|\sqrt{{\left({x}_{1}\right)}^{2}+{\left({y}_{1}\right)}^{2}}-r\text{\hspace{0.17em}}\right|$ .

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 $\left(h,k\right)$ and a radius $r$ , the shortest distance between the point $P\left({x}_{1},{y}_{1}\right)$ and the circle is $\left|\sqrt{{\left({x}_{1}-h\right)}^{2}+{\left({y}_{1}-k\right)}^{2}}-r\text{\hspace{0.17em}}\right|$ .

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Example 1:
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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 $3$ .

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

$\begin{array}{l}D=\left|\text{\hspace{0.17em}}\sqrt{{\left(3\right)}^{2}+{\left(4\right)}^{2}}-3\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\text{\hspace{0.17em}}\sqrt{25}-3\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\text{\hspace{0.17em}}5-3\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\end{array}$

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

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Example 2:
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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 $6$ .

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

$\begin{array}{l}D=\left|\sqrt{{\text{\hspace{0.17em}}(-2)}^{2}+{\left(2\right)}^{2}}-6\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\text{\hspace{0.17em}}\sqrt{8}-6\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6-2\sqrt{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx 3.17\end{array}$

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

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

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

${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$ 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.

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

$\begin{array}{l}D=\left|\text{\hspace{0.17em}}5-\sqrt{{\left(-3-\left(-2\right)\right)}^{2}+{\left(3-0\right)}^{2}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\text{\hspace{0.17em}}5-\sqrt{1+9}\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\text{\hspace{0.17em}}5-\sqrt{10}\text{\hspace{0.17em}}\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx 1.84\end{array}$

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

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

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

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

So, the circle has its center at $(4,-5)$ and has a radius of $7$ units.

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

$\begin{array}{c}D=\left|\text{\hspace{0.17em}}\sqrt{{\left(-4-4\right)}^{2}+{\left(-11-\left(-5\right)\right)}^{2}}-7\text{\hspace{0.17em}}\right|\\ =\left|\text{\hspace{0.17em}}\sqrt{64+36}-7\text{\hspace{0.17em}}\right|\\ =\sqrt{100}-7\\ =3\end{array}$

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