Answer:
The equation of circle with center (-2, 1) and radius 3 is [tex]x^2+4x+y^2-2y=4[/tex]
Step-by-step explanation:
Given a circle with center (-2, 1) and radius 3.
We have to find its equation.
The standard form for equation of a circle with center (h,k) and radius r is given by, [tex](x-h)^2+(y-k)^2=r^2[/tex]
For given circle with center (-2, 1) and radius 3.
h = -2 , k = 1 , and r = 3
Substitute, above , we have,
[tex](x+2)^2+(y-1)^2=3^2[/tex]
Solving ,
Using algebraic identity,
[tex](a+b)^2=a^2+b^2+2ab\\ (a-b)^2=a^2+b^2-2ab[/tex]
we get,
[tex]x^2+4+4x+y^2+1-2y=9[/tex]
Simplify, we get,
[tex]x^2+4x+y^2-2y+5=9[/tex]
subtract 5 both side,
[tex]x^2+4x+y^2-2y=4[/tex]
Thus, The equation of circle with center (-2, 1) and radius 3 is [tex]x^2+4x+y^2-2y=4[/tex]
