# What is the equation for the circle which passes through (3;4) (-4;3) (5;0)?

Jul 19, 2018

I will start you off

#### Explanation:

$\textcolor{b l u e}{\text{Building the initial condition model}}$

Let some constant related to $x$ be ${K}_{x}$
Let some constant related to $y$ be ${K}_{y}$

Let the radius be $r$

$\textcolor{g r e e n}{\text{Then as a general equation we have:}}$

$\textcolor{g r e e n}{{\left(x - {K}_{x}\right)}^{2} + {\left(y - {K}_{y}\right)}^{2} = {r}^{2} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)}$
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For point 1 $\to {P}_{1} \to \left(- 4 , 3\right) \text{ } E q u a t i o n \left(1\right)$ becomes:

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

${\left({K}_{x}\right)}^{2} + 8 {K}_{x} + 16 + {\left({K}_{y}\right)}^{2} - 6 {K}_{y} + 9 = {r}^{2}$

$\textcolor{b r o w n}{{\left({K}_{x}\right)}^{2} + {\left({K}_{y}\right)}^{2} + 8 {K}_{x} - 6 {K}_{y} + 25 = {r}^{2} \textcolor{w h i t e}{\text{ddd}} E q n \left({1}_{a}\right)}$
......................................................................................

For point 2 $\to {P}_{2} \to \left(3 , 4\right) \text{ } E q u a t i o n \left(1\right)$ becomes:

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

${\left({K}_{x}\right)}^{2} - 6 {K}_{x} + 9 + {\left({K}_{y}\right)}^{2} - 8 {K}_{y} + 16 = {r}^{2}$

$\textcolor{b r o w n}{{\left({K}_{x}\right)}^{2} + {\left({K}_{y}\right)}^{2} - 6 {K}_{x} - 8 {K}_{y} + 25 = {r}^{2} \textcolor{w h i t e}{\text{ddd}} E q n \left({1}_{b}\right)}$

...........................................................................
For point 3 $\to {P}_{2} \to \left(3 , 4\right) \text{ } E q u a t i o n \left(1\right)$ becomes:

${\left(5 - {K}_{x}\right)}^{2} + {\left(0 - {K}_{y}\right)}^{2} = {r}^{2}$

${\left({K}_{x}\right)}^{2} - 10 {K}_{x} + 25 + {\left({K}_{y}\right)}^{2} = {r}^{2}$

color(brown)((K_x)^2+(K_y)^2-10K_x+25=r^2color(white)("ddd") Eqn(1_c)
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$\textcolor{b l u e}{\text{Putting it all together}}$

$\textcolor{b r o w n}{{\left({K}_{x}\right)}^{2} + {\left({K}_{y}\right)}^{2} + 8 {K}_{x} - 6 {K}_{y} + 25 = {r}^{2} \textcolor{w h i t e}{\text{ddd}} E q n \left({1}_{a}\right)}$
$\textcolor{b r o w n}{{\left({K}_{x}\right)}^{2} + {\left({K}_{y}\right)}^{2} - 6 {K}_{x} - 8 {K}_{y} + 25 = {r}^{2} \textcolor{w h i t e}{\text{ddd}} E q n \left({1}_{b}\right)}$
color(brown)((K_x)^2+(K_y)^2-10K_xcolor(white)("ddddd")+25=r^2color(white)("ddd") Eqn(1_c)

Now you manipulate these to gradually isolate the unknowns.
3 unknowns and 3 equations so solvable.