Question

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

Answer 1

`sqrt(10)-sqrt(5) ~~ 0.9262`

Explanation:

`0 = x^2-6x+y^2+2y`

`= (x-3)^2+(y-(-1))^2-10`

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

`(x-3)^2+(y-(-1))^2 = (sqrt(10))^2`

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

`(x-h)^2+(y-k)^2 = r^2`

with centre `(h, k) = (3, -1)` and radius `r = sqrt(10)`

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

`sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((3-2)^2+(-1-1)^2) = sqrt(1+4) = sqrt(5) < sqrt(10)`

So the point `(2, 1)` 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]}

Answer 2

`d=0.92621`

Explanation:

Given a circle

`C->(x-x_c)^2+(y-y_c)^2 = r^2`

and a point `q_0=(x_0,y_0)`, the minimum distance from `q_0` to the circle `C` is given by

`d = norm(q_0-p_0)`

where `p_0` is the intersection of `C` and the straight

`S->s = q_0+lambda(p_c-q_0)`

The `C` equation can be written as `C->(p-p_c).(p-p_c) = r^2`

with `p = (x,y)`. In the intersection occurs `p = s` then substituting

`C nn S ->(q_0+lambda(p_c-q_0)).(q_0+lambda(p_c-q_0))=r^2`

after simplifications we get at

`lambda^2-2lambda+1=r^2/((p_c-q_0).(p_c-q_0))`

or

`(lambda-1)^2=r^2/((p_c-q_0).(p_c-q_0))`

or

`lambda = 1pmr/sqrt((p_c-q_0).(p_c-q_0))`

Putting numeric values we get at

`lambda = 1 pm sqrt(2)`

and after substitution in `S`

`p_0^1=(2 sqrt[2] + 3 (1 - sqrt[2]), -1 + 2 sqrt[2])`
`p_0^2 = (-2 sqrt[2] + 3 (1 + sqrt[2]), -1 - 2 sqrt[2])`

and the nearest is `q_0^1` giving a distance of

`norm(p_0^1-q_0)=0.92621`