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

QUESTION

# 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)