# How do you write an equation for a circle with center in quadrant two, Radius 3, Tangent to y-axis at (0,4)?

Nov 15, 2016

${x}^{2} + {y}^{2} + 6 x - 8 y + 16 = 0$

#### Explanation:

As it is tangent to $y$-axis at $\left(0 , 4\right)$, center is in second quadrant and radius is $3$,

the center is $3$ units to the left of $\left(0 , 4\right)$ i.e. center is $\left(- 3 , 4\right)$

Hence equation of circle is

${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 4\right)}^{2} = 9$

i.e. ${\left(x + 3\right)}^{2} + {\left(y - 4\right)}^{2} = 9$

or ${x}^{2} + 6 x + 9 + {y}^{2} - 8 y + 16 = 9$

or ${x}^{2} + {y}^{2} + 6 x - 8 y + 16 = 0$
graph{x^2+y^2+6x-8y+16=0 [-8.78, 4.56, -0.8, 5.9]}