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]}