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

Mar 17, 2018

$\left(3 , - 4\right)$

#### Explanation:

the standard eqn of a circle with centre $\left(a , b\right) \mathmr{and} r a \mathrm{di} u s$r

is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2} - - \left(1\right)$

so for

${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = 25$

comparing this with $\left(1\right)$

the centre will be

$\left(3 , - 4\right)$

for good measure the radius will be

$r = \sqrt{25} = 5$

Mar 17, 2018

Centre $\to \left(x , y\right) = \left(+ 3 , - 4\right)$

#### 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 $\left(x , y\right) = \left(+ 3 , - 4\right)$

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Consider the ${\left(\textcolor{red}{x} \textcolor{b l u e}{- 3}\right)}^{2}$

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

So color(red)(x_(centre")=+3) because $\textcolor{red}{+ 3} \textcolor{b l u e}{- 3} = 0 \leftarrow \text{ the origin}$

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Consider the ${\left(\textcolor{red}{y} \textcolor{b l u e}{+ 4}\right)}^{2}$

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

So color(red)(y_(centre")=-4)# because $\textcolor{red}{- 4} \textcolor{b l u e}{+ 4} = 0 \leftarrow \text{ the origin}$
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