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# What is the standard form of the equation of a circle with center (-3,3) and tangent to the line y=1?

Equation of circle is ##x^2+y^2+6x-6y+14=0## and ##y=1## is tangent at ##(-3,1)##

The equation of a circle with center ##(-3,3)## with radius ##r## is

##(x+3)^2+(y-3)^2=r^2##

or ##x^2+y^2+6x-6y+9+9-r^2=0##

As ##y=1## is a tangent to this circle, putting ##y=1## in the equation of a circle should give only one solution for ##x##. Doing so we get

##x^2+1+6x-6+9+9-r^2=0## or

##x^2+6x+13-r^2=0##

and as we should have only one solution, discriminant of this quadratic equation should be ##0##.

Hence, ##6^2-4xx1xx(13-r^2)=0## or

##36-52+4r^2=0## or ##4r^2=16## and as ##r## has to be positive

##r=2## and hence equation of circle is

##x^2+y^2+6x-6y+9+9-4=0## or ##x^2+y^2+6x-6y+14=0##

and ##y=1## is tangent at ##(-3,1)##