Question

The circle below is centered at the point (2,-1) and has a radius of length 3. What is iis equation?

Answer 1

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{2}{ h},\stackrel{-1}{ k})\qquad \qquad radius=\stackrel{3}{ r}\\[2em] [x-2]^2+[y-(-1)]^2=3^2\implies (x-2)^2+(y+1)^2=9$

Answer 2

Answer: $(x-2)^2+(y+1)^2=9$

Step-by-step explanation:

The equation of a circle having center (h,k) and radius r is given by :-

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

Given: Vertex of circle= (2,-1)

Radius of circle= 3 units

Then the equation of a circle will be :

$(x-2)^2+(y-(-1))^2=3^2\\\\\Rightarrow\ (x-2)^2+(y+1)^2=9$

Hence, the equation of circle = $(x-2)^2+(y+1)^2=9$