# How do you write an equation for a circle with diameter has endpoints(-3,4) and (4,-3)?

May 23, 2016

The equation of the circle is:

$\textcolor{b r o w n}{\implies \frac{49}{2} = {\left({x}_{i} - \frac{1}{2}\right)}^{2} + {\left({y}_{i} - \frac{1}{2}\right)}^{2}}$

#### Explanation:

The centre will be at the middle of these points so it is the mean value.

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(- 3 , 4\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(4 , - 3\right)$
Let the centre be ${P}_{c} \to \left({x}_{x} , {y}_{c}\right)$
Let the radius be $r$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the centre}}$

Mean for x$\to \frac{4 + \left(- 3\right)}{2} = \frac{1}{2} \to {x}_{c}$

Mean for y$\to \frac{4 + \left(- 3\right)}{2} = \frac{1}{2} \to {y}_{c}$

$\textcolor{b l u e}{\text{Centre } \to \left({x}_{c} , {y}_{c}\right) \to \left(\frac{1}{2} , \frac{1}{2}\right)}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the magnitude of the radius}}$
Using Pythagoras

I chose $\left({x}_{1} , {y}_{1}\right)$ to determine the length of $r$

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

${r}^{2} = {\left[\frac{1}{2} - \left(- 3\right)\right]}^{2} + {\left[\frac{1}{2} - 4\right]}^{2}$

$r = \sqrt{\frac{49}{4} + \frac{49}{4}} = \sqrt{\frac{49}{2}}$

$r = \frac{7}{\sqrt{2}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{2}}{\sqrt{2}}$

$\textcolor{b l u e}{r = \frac{7 \sqrt{2}}{2} \text{ "->" } {r}^{2} = \frac{49}{2}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

Let any $x$ be ${x}_{i}$
Let any $y$ be ${y}_{i}$

The equation of this circle is

${r}^{2} = {\left({x}_{i} - {x}_{x}\right)}^{2} + {\left({y}_{i} - {y}_{c}\right)}^{2}$

$\textcolor{b l u e}{\implies \frac{49}{2} = {\left({x}_{i} - \frac{1}{2}\right)}^{2} + {\left({y}_{i} - \frac{1}{2}\right)}^{2}}$