How do you write the equation for a circle where the points passes through the points (1,1), (-2, 2), and (-5,1)?

Nov 13, 2016

${\left(x - - 2\right)}^{2} + {\left(y - - 3\right)}^{2} = {5}^{2}$

Explanation:

Use the standard form for the equation of a circle:

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

And the points, $\left(1 , 1\right) , \left(- 2 , 2\right) , \mathmr{and} \left(- 5 , 1\right)$

To write 3 equations:

${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {r}^{2} \text{ [1]}$
${\left(- 2 - h\right)}^{2} + {\left(2 - k\right)}^{2} = {r}^{2} \text{ [2]}$
${\left(- 5 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {r}^{2} \text{ [3]}$

Temporarily eliminate the variable r by setting the left side of equation [1] equal to the left side of equation [2] and the same for equation [1] and equation [3]:

${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {\left(- 2 - h\right)}^{2} + {\left(2 - k\right)}^{2}$
${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {\left(- 5 - h\right)}^{2} + {\left(1 - k\right)}^{2} \text{ [4]}$

Please notice that equation [4] has ${\left(1 - k\right)}^{2}$ on both sides so we can subtract this form both sides:

${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {\left(- 2 - h\right)}^{2} + {\left(2 - k\right)}^{2}$
${\left(1 - h\right)}^{2} = {\left(- 5 - h\right)}^{2}$

Use the pattern ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$ to expand all of the squares:

$1 - 2 h + {h}^{2} + 1 - 2 k + {k}^{2} = 4 + 4 h + {h}^{2} + 4 - 4 k + {k}^{2}$
$1 - 2 h + {h}^{2} = 25 + 10 h + {h}^{2}$

The ${h}^{2} \mathmr{and} {k}^{2}$ terms cancel:

$1 - 2 h + 1 - 2 k = 4 + 4 h + 4 - 4 k \text{ [5]}$
$1 - 2 h = 25 + 10 h \text{ [6]}$

Use equation 6 to solve for h:

$- 12 h = 24$

$h = - 2$

Combine like tems of equation [5] with k terms on the left and constant and h terms on the right:

$2 k = 6 h + 6$

$k = 3 h + 3$

Substitute -2 for h:

$k = 3 \left(- 2\right) + 3$

$k = - 3$

Substitute the values for h and k into one of the first three equations (I will use equation [2]):

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

${\left(0\right)}^{2} + {\left(5\right)}^{2} = {r}^{2}$

Solve for r:

$r = 5$

The equation of the circle is:

${\left(x - - 2\right)}^{2} + {\left(y - - 3\right)}^{2} = {5}^{2}$