# What is the standard form of the equation of a circle passing through (0, -14), (-12, -14), and (0,0)?

Jan 14, 2017

A circle of radius $\sqrt{85}$ and centre $\left(- 6 , - 7\right)$

Standard form equation is: ${\left(x + 6\right)}^{2} + {\left(y + 7\right)}^{2} = 85$

Or, ${x}^{2} + 12 x + {y}^{2} + 14 y = 0$

#### Explanation:

The Cartesian equation of a circle with centre $\left(a , b\right)$ and radius $r$ is:

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

If the circle passes through (0,-14) then:

${\left(0 - a\right)}^{2} + {\left(- 14 - b\right)}^{2} = {r}^{2}$
${a}^{2} + {\left(14 + b\right)}^{2} = {r}^{2}$ ................................ [1]

If the circle passes through (0,-14) then:

${\left(- 12 - a\right)}^{2} + {\left(- 14 - b\right)}^{2} = {r}^{2}$
${\left(12 + a\right)}^{2} + {\left(14 + b\right)}^{2} = {r}^{2}$ ................................ [2]

If the circle passes through (0,0) then:

${\left(0 - a\right)}^{2} + {\left(0 - b\right)}^{2} = {r}^{2}$
${a}^{2} + {b}^{2} = {r}^{2}$ ................................ [3]

We now have 3 equations in 3 unknowns

Eq [2] - Eq [1] gives:

${\left(12 + a\right)}^{2} - {a}^{2} = 0$
$\therefore \left(12 + a - a\right) \left(12 + a + a\right) = 0$
$\therefore 12 \left(12 + 2 a\right) = 0$
$\therefore a = - 6$

Subs $a = 6$ into Eq [3]:

$36 + {b}^{2} = {r}^{2}$ ................................ [4]

Subs $a = 6$ and ${r}^{2} = 36 + {b}^{2}$into Eq [1]:

$36 + {\left(14 + b\right)}^{2} = 36 + {b}^{2}$
$\therefore {\left(14 + b\right)}^{2} - {b}^{2} = 0$
$\therefore \left(14 + b - b\right) \left(14 + b + b\right) = 0$
$\therefore 14 \left(14 + 2 b\right) = 0$
$\therefore b = - 7$

And finally, Subs $b = - 7$ into Eq [4];

$36 + 49 = {r}^{2}$
$\therefore {r}^{2} = 85$
$\therefore r = \sqrt{85}$

And so the equation of the circle is

${\left(x + 6\right)}^{2} + {\left(y + 7\right)}^{2} = 85$

Which represents a circle of radius $\sqrt{85}$ and centre $\left(- 6 , - 7\right)$

We can multiply out if required to get:

${x}^{2} + 12 x + 36 + {y}^{2} + 14 y + 49 = 85$
${x}^{2} + 12 x + {y}^{2} + 14 y = 0$