Question

The shortest distance between the circle

Answer 1

\[x^2-6x+y^2-8y+21=0\] and the origin is

Answer 2

The first thing I'd do is complete the square for the x terms. Do you know how to do that?

Answer 3

Yeah. I got \[(x-3)^2+(y-4)^2=4\]

Answer 4

Is this correct? @jim_thompson5910

Answer 5

let me check

Answer 6

good, you also took care of the y terms too

ok the next step is to solve for y

Answer 7

how would I do that? divide by 4?

Answer 8

@jim_thompson5910

Answer 9

hmmm this is looking to be more complicated than I thought. Let me see if I can find a quicker way

Answer 10

ok it's a bit complicated but let's see how things go

anyways, the next step is to solve for y, so let's do that


\[\large (x-3)^2+(y-4)^2=4\]

\[\large (y-4)^2=4-(x-3)^2\]

\[\large y-4=\pm\sqrt{4-(x-3)^2}\]

\[\large y=4\pm\sqrt{4-(x-3)^2}\]


hopefully you agree with the steps so far?

Answer 11

Yeah I get how you did that.

Answer 12

And then how would I find the shortest distance between the circle and the origin?

Answer 13

let's plot point P on the circle

P is in the form (x,y)

it turns out that y breaks down into these two equations

\[\large y=4+\sqrt{4-(x-3)^2}\]
\[\large y=4-\sqrt{4-(x-3)^2}\]

Answer 14

you'll have to use the distance formula to find the distance from (0,0) which is the origin to P(x,y)

Answer 15

So what is P(x,y)?

Answer 16

we don't know x yet, so we leave it as x

replace y with the second equation that I wrote down. You'll use the second equation because the first one is too far. So I'll save you that bit of work

Answer 17

Recall that the distance formula is this

\[\large d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

Let's perform substitutions/replacements to get...

\[\large d = \sqrt{\left({\color{red}{x_2}}-{\color{green}{x_1}}\right)^2 + \left({\color{blue}{y_2}}-{\color{purple}{y_1}}\right)^2}\]

\[\large d = \sqrt{\left({\color{red}{x}}-{\color{green}{0}}\right)^2 + \left({\color{blue}{4-\sqrt{4-(x-3)^2}}}-{\color{purple}{0}}\right)^2}\]

\[\large d = \sqrt{x^2 + \left(4-\sqrt{4-(x-3)^2}\right)^2}\]

with me so far?

Answer 18

Yeah. And then how would I solve that?

Answer 19

we want to minimize this distance d, so we need to find the derivative of this function and set it equal to zero

Answer 20

I'm assuming this is a calculus class?

Answer 21

Very close guess! It's actually precalculus

Answer 22

ok so forget about the derivative then since you haven't gotten to that part yet

Answer 23

you'll need to use a graphing calculator

plot this function

\[\large y = \sqrt{x^2 + \left(4-\sqrt{4-(x-3)^2}\right)^2}\]

Answer 24

then use the `minimize` feature to find the local min. This will lead to the smallest distance and the x value needed to get there

Answer 25

I got (2.5,2.75)

Answer 26

incorrect. Make sure you typed in the function correctly

Answer 27

When I graph it, it gives me a weird shape. It's suppose to be a circle right?

Answer 28

no, you'll get some half-ellipse

Answer 29

this is what you should get when you plot that d function (see attached)

Answer 30

oh okay so the vertex is (2,3) and so then I would do the distance formula again and use (2,3) and (0,0) right?

Answer 31

the vertex isn't (2,3)
the lowest point on that red curve I posted is (1.8, 3)

so when x = 1.8 the distance is minimized to d = 3

plug x = 1.8 into the y function to find the value of y

Answer 32

Okay I got y=2.4

Answer 33

me too, so (1.8, 2.4) is the point that is closest to (0,0)

that distance is 3 units

Answer 34

any other point on this circle will have a distance larger than 3 units

Answer 35

oh okay thank you so much!! :)

Answer 36

you're welcome