# What is the standard form of the equation of a circle passing through (0,8), (5,3) and (4,6)?

Nov 29, 2016

I have taken you to a point where you should be able to take over.

#### Explanation:

$\textcolor{red}{\text{There may be an easier way of doing this}}$

The trick is to manipulate these 3 equations in such a way that you end up with 1 equation with 1 unknown.

Consider the standard form of ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(0 , 8\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(5 , 3\right)$
Let point 3 be ${P}_{3} \to \left({x}_{3} , {y}_{3}\right) = \left(4 , 6\right)$
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For ${P}_{1} \to {\left({x}_{1} - a\right)}^{2} + {\left({y}_{1} - b\right)}^{2} = {r}^{2}$

${\left(0 - a\right)}^{2} + {\left(8 - b\right)}^{2} = {r}^{2}$

${a}^{2} + 64 - 16 b + {b}^{2} = {r}^{2}$...............Equation(1)
............................................................................................................
For ${P}_{2} \to {\left({x}_{2} - a\right)}^{2} + {\left({y}_{2} - b\right)}^{2} = {r}^{2}$

${\left(5 - a\right)}^{2} + {\left(3 - b\right)}^{2} = {r}^{2}$

$25 - 10 a + {a}^{2} + 9 - 6 b + {b}^{2} = {r}^{2}$

${a}^{2} - 10 a + 34 - 6 b + {b}^{2} = {r}^{2}$............Equation(2)
..........................................................................................................

For ${P}_{3} \to {\left({x}_{3} - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

${\left(4 - a\right)}^{2} + {\left(6 - b\right)}^{2} = {r}^{2}$

$16 - 8 a + {a}^{2} + 36 - 12 b + {b}^{2} = {r}^{2}$

${a}^{2} - 8 a + 52 - 12 b + {b}^{2} = {r}^{2}$...........Equation(3)
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Lets see where this gets us!

Equation(3) - Equation(2)

${a}^{2} - 8 a - 12 b + {b}^{2} + 52 = {r}^{2}$
$\underline{{a}^{2} - 10 a - 6 b + {b}^{2} + 34 = {r}^{2}} \leftarrow \text{ subtract}$
$0 \text{ } + 2 a - 6 b + 0 + 18 = 0$

$2 a - 6 b + 18 = 0$ ...........................Equation(4)

$\implies a = \frac{6 b - 18}{2} = 3 b - 9$

$\textcolor{b r o w n}{\text{we can now substitute for } a}$$\textcolor{b r o w n}{\text{in equations 1 and 2 and solve for } b}$

$e q u a t i o n \left(1\right) = {r}^{2} = e q u a t i o n \left(2\right)$

${a}^{2} - 16 b + {b}^{2} \text{ " =" } {a}^{2} - 10 a - 6 b + {b}^{2} + 34$

$\cancel{{a}^{2}} - 16 b + \cancel{{b}^{2}} \text{ " =" } \cancel{{a}^{2}} - 10 a - 6 b + \cancel{{b}^{2}} + 34$

Substituting for $a$

$- 16 b \text{ "=" } - 10 \left(3 b - 9\right) - 6 b + 34$

$- 16 b \text{ "=" } - 30 b + 90 - 6 b + 34$

$- 16 b \text{ "=" } - 36 b + 124$

" "color(green)(ul(bar(|" "b=124/20=31/5" "|))

$\textcolor{red}{\text{I will let you take it on from this point}}$