Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
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)##