What is the equation of a circle with center #(2, -3)# and radius of #3#?

2 Answers
Ratnaker Mehta
Jul 25, 2016

#x^2+y^2-4x+6y+4=0#.

Explanation:

The eqn. of a circle with centre #(h,k)# and radius #r# is given by,

#(x-h)^2+(y-k)^2=r^2#.

Accordingly, the reqd. eqn. is,

#(x-2)^2+(y+3)^2=3^2#, i.e.,

#x^2+y^2-4x+6y+4=0#.

Hitesh C Nimbalkar
Jul 25, 2016

Hi.
Equation of standard circle i.e circle having its centre at origin is
#x^2#+#y^2#=#r^2#
here r is the radius
Now if the circle does not have its centre at origin
then
if it is having its cente at poin (X1,Y1)
then
equation of circle can be written as
#(x-X1)^2#+#(y-Y1)^2#=#r^2#
then for now for above question points are (2,-3)
and radius is 3
then equation is
#(x-2)^2#+#(y-(-3))^2#=9

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