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

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Answer 1 · 2016-11-15T10:50:31

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

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

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

Hence equation of circle is

$(x-(-3))^2+(y-4)^2=9$

i.e. $(x+3)^2+(y-4)^2=9$

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

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