Question

How do you find the center of the circle with equation `(x - 3)^ 2 + (y + 4)^ 2 = 25`?

Answer 1

`(3,-4)`

Explanation:

the standard eqn of a circle with centre `(a,b) and radius`r#

is

`(x-a)^2+(y-b)^2=r^2--(1)`

so for

`(x-3)^2+(y+4)^2=25`

comparing this with `(1)`

the centre will be

`(3,-4)`

for good measure the radius will be

`r=sqrt25=5`

Answer 2

Centre `->(x,y)=(+3,-4)`

Explanation:

Note that the equation of a circle centred at the origin is derived using Pythagoras. This is because you can form a triangle related to any point on the perimeter. `x^2+y^2=r^2` where `r` is the radius.

The behaviour of the question's equation is such that it models the following:

Draw the circle at the origin where `r` is of the correct magnitude.
Move every point that is at `x-3` and plot it where `x` is.
Move every point that is at `y+4` and plot it where `y` is

The net effect is that you move the plot to the right by 3 and down by 4. This puts the new centre at `(x,y)=(+3,-4)`

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Consider the `(color(red)(x)color(blue)(-3))^2`

This is in fact: `x_"centre"-3=x_("origin")=0`

So `color(red)(x_(centre")=+3)` because `color(red)(+3)color(blue)(-3)=0larr" the origin"`

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Consider the `(color(red)(y)color(blue)(+4))^2`

This is in fact: `y_"centre"+4=y_("origin")=0`

So `color(red)(y_(centre")=-4)` because `color(red)(-4)color(blue)(+4)=0larr" the origin"`
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