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

1 Answer
May 7, 2016

Answer:

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

Explanation:

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)#