[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{2}{ h},\stackrel{-1}{ k})\qquad \qquad radius=\stackrel{3}{ r}\\[2em] [x-2]^2+[y-(-1)]^2=3^2\implies (x-2)^2+(y+1)^2=9[/tex]
Answer: [tex](x-2)^2+(y+1)^2=9[/tex]
Step-by-step explanation:
The equation of a circle having center (h,k) and radius r is given by :-
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Given: Vertex of circle= (2,-1)
Radius of circle= 3 units
Then the equation of a circle will be :
[tex](x-2)^2+(y-(-1))^2=3^2\\\\\Rightarrow\ (x-2)^2+(y+1)^2=9[/tex]
Hence, the equation of circle = [tex](x-2)^2+(y+1)^2=9[/tex]
