How do you find general form of circle with Center at the point (-3, 1) and tangent to the y-axis?

Dec 9, 2015

${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = {3}^{2}$

Explanation:

${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = {r}^{2}$

y-axis: $x = 0$

${3}^{2} + {\left(y - 1\right)}^{2} = {r}^{2}$

${\left(y - 1\right)}^{2} = {r}^{2} - 9$

y = 1 ± sqrt {r^2 - 9}

y is unique only if ${r}^{2} - 9 = 0 R i g h t a r r o w r = 3$