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`

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`