# How do you find the distance from the point (2.1) to the circle x^2 - 6x + y^2 + 2y= 0?

May 30, 2016

$\sqrt{10} - \sqrt{5} \approx 0.9262$

#### Explanation:

$0 = {x}^{2} - 6 x + {y}^{2} + 2 y$

$= {\left(x - 3\right)}^{2} + {\left(y - \left(- 1\right)\right)}^{2} - 10$

Adding $10$ to both sides and transposing, we can express this as:

${\left(x - 3\right)}^{2} + {\left(y - \left(- 1\right)\right)}^{2} = {\left(\sqrt{10}\right)}^{2}$

which is the standard form of the equation of a circle:

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

with centre $\left(h , k\right) = \left(3 , - 1\right)$ and radius $r = \sqrt{10}$

The distance between the point $\left(2 , 1\right)$ and the centre of the circle $\left(3 , - 1\right)$ is given by the distance formula:

$\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$= \sqrt{{\left(3 - 2\right)}^{2} + {\left(- 1 - 1\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5} < \sqrt{10}$

So the point $\left(2 , 1\right)$ is inside the circle at a distance of $\sqrt{10} - \sqrt{5}$ from the circumference.

graph{(x^2-6x+y^2+2y)((x-2)^2+(y-1)^2-0.02)((x-3)^2+(y+1)^2-0.02) = 0 [-10, 10, -5, 5]}

May 30, 2016

$d = 0.92621$

#### Explanation:

Given a circle

$C \to {\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$

and a point ${q}_{0} = \left({x}_{0} , {y}_{0}\right)$, the minimum distance from ${q}_{0}$ to the circle $C$ is given by

$d = \left\lVert {q}_{0} - {p}_{0} \right\rVert$

where ${p}_{0}$ is the intersection of $C$ and the straight

$S \to s = {q}_{0} + \lambda \left({p}_{c} - {q}_{0}\right)$

The $C$ equation can be written as $C \to \left(p - {p}_{c}\right) . \left(p - {p}_{c}\right) = {r}^{2}$

with $p = \left(x , y\right)$. In the intersection occurs $p = s$ then substituting

$C \cap S \to \left({q}_{0} + \lambda \left({p}_{c} - {q}_{0}\right)\right) . \left({q}_{0} + \lambda \left({p}_{c} - {q}_{0}\right)\right) = {r}^{2}$

after simplifications we get at

${\lambda}^{2} - 2 \lambda + 1 = {r}^{2} / \left(\left({p}_{c} - {q}_{0}\right) . \left({p}_{c} - {q}_{0}\right)\right)$

or

${\left(\lambda - 1\right)}^{2} = {r}^{2} / \left(\left({p}_{c} - {q}_{0}\right) . \left({p}_{c} - {q}_{0}\right)\right)$

or

$\lambda = 1 \pm \frac{r}{\sqrt{\left({p}_{c} - {q}_{0}\right) . \left({p}_{c} - {q}_{0}\right)}}$

Putting numeric values we get at

$\lambda = 1 \pm \sqrt{2}$

and after substitution in $S$

${p}_{0}^{1} = \left(2 \sqrt{2} + 3 \left(1 - \sqrt{2}\right) , - 1 + 2 \sqrt{2}\right)$
${p}_{0}^{2} = \left(- 2 \sqrt{2} + 3 \left(1 + \sqrt{2}\right) , - 1 - 2 \sqrt{2}\right)$

and the nearest is ${q}_{0}^{1}$ giving a distance of

$\left\lVert {p}_{0}^{1} - {q}_{0} \right\rVert = 0.92621$