How do you write an equation for a circle tangent to the y-axis and with the center (5, 6)?

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Answer 1 · 2016-09-01T05:02:07

# x^2+y^2-10x-12y+36=0#

The reqd. circle touches the Y-axis [ eqn. #x=0#] and has the Centre

at #C(5,6)#.

From Geometry , we know that, in the given situation, the #bot#-

distance from the Centre to the tgt. must equal the

Radius #r# of the circle.

#:. r=|5|=5#.

With Centre #(5,6) and r=5#, the eqn. of circle is,

# (x-5)^2+(y-6)^2=5^2-12y+36=0#. i.e.,

graph{x^2+y^2-10x-12y+36=0 [-22.8, 22.8, -11.4, 11.42]} #x^2+y^2-10x-12y+36=0#