pre calc

Question

pre calc

Mathplayer223 · Answer

attachment

Mathplayer223 · Answer

@snowflake0531

snowflake0531 · Answer

This one is relatively easy, just use the form $(x-h)^2 + (y-k)^2 = r^2$ and plug h k and r in

snowflake0531 · Answer

they gave you all the numbers already

Mathplayer223 · Answer

(x-1)^2+(y-k)^2=10^2

snowflake0531 · Answer

it's 1 over 10, not 10
and it's ( 1/10 , 0\) 1/10 is x, 0 is y

Mathplayer223 · Answer

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

Mathplayer223 · Answer

@vocaloid

Mathplayer223 · Answer

attachment

Mathplayer223 · Answer

i solved the first half

AZ · Answer

@mathplayer223 wrote:

i solved the first half

You finally got the first half, well done! To find the general form of the equation, you need to simplify it hint: expand $ \left(x - \dfrac{1}{10}\right)^2$ use the rule $ (a-b)^2 = a^2 -2ab + b^2$

Mathplayer223 · Answer

x^2-x/5+1/100

AZ · Answer

Good!

You just went from

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

$x^2 - \dfrac{x}{5} + \dfrac{1}{100} + y^2 = \left(\dfrac{1}{10}\right)^2$

Can you simplify the right hand side? the (1/10)^2 and then subtract it on both sides?

Mathplayer223 · Answer

am i solving for x or do i find the center

Mathplayer223 · Answer

is there a square root in the answer?

AZ · Answer

@mathplayer223 wrote:

am i solving for x or do i find the center

neither, we're just rearranging the equation so that way it equals to 0 and it would be in the general form instead of the standard form which is $(x-h)^2 + (y-k)^2 = r^2$ we can't solve for 'x' because to solve for it, we need another equation since we have two variables (x and y) in this one equation. When you have two variables, you need two equations to solve for one. If we wanted to find the center, we would have used the original equation and not expanded it

AZ · Answer

@mathplayer223 wrote:

is there a square root in the answer?

There shouldn't be First, what is (1/10)^2 = ? Hint: (1/10) * (1/10) = ?

AZ · Answer

@az wrote:

neither, we're just rearranging the equation so that way it equals to 0 and it would be in the general form instead of the standard form which is $(x-h)^2 + (y-k)^2 = r^2$ we can't solve for 'x' because to solve for it, we need another equation since we have two variables (x and y) in this one equation. When you have two variables, you need two equations to solve for one. If we wanted to find the center, we would have used the original equation and not expanded it

to clarify, the standard form equation for a circle is $(x-h)^2 + (y-k)^2 = r^2$ the general form equation is when you expand and simplify and set the equation equal to 0

Mathplayer223 · Answer

(x-1/10)^2+y^2=1/100

AZ · Answer

@mathplayer223 wrote:

(x-1/10)^2+y^2=1/100

Yes, it's 1/100 and like you said earlier (x - 1/10)^2 = x^2-x/5+1/100 so we can write that altogether as x^2-x/5+1/100 + y^2 = 1/100 can you see that you can subtract 1/100 on both sides now?

Mathplayer223 · Answer

yes

AZ · Answer

once you simplify it by subtracting 1/100 on both sides, you're all done because the equation is set equal to 0

Mathplayer223 · Answer

how do i subtract by both sides

AZ · Answer

for example if you had

a + 10 = 10

you would subtract 10 on both sides

a + 10 - 10 = 10 - 10
a = 0

you would see that it cancels out since you have the same term on both sides

Mathplayer223 · Answer

x^2-x/5+0 + y^2 = 0

AZ · Answer

there you go!

and adding 0 means nothing so you can just write that as

x^2 - x/5 + y^2 = 0

Mathplayer223 · Answer

thanks

AZ · Answer

you're welcome!

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