What is the equation of a circle with center (2, -3)(2,−3) and radius of 33?

2 Answers
Ratnaker Mehta
Jul 25, 2016

x^2+y^2-4x+6y+4=0x2+y2−4x+6y+4=0.

Explanation:

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

(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2.

Accordingly, the reqd. eqn. is,

(x-2)^2+(y+3)^2=3^2(x−2)2+(y+3)2=32, i.e.,

x^2+y^2-4x+6y+4=0x2+y2−4x+6y+4=0.

Hitesh C Nimbalkar
Jul 25, 2016

Hi.
Equation of standard circle i.e circle having its centre at origin is
x^2x2+y^2y2=r^2r2
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(x−X1)2+(y-Y1)^2(y−Y1)2=r^2r2
then for now for above question points are (2,-3)
and radius is 3
then equation is
(x-2)^2(x−2)2+(y-(-3))^2(y−(−3))2=9

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