Question

Enter the equation of the circle described below.

Center (-1, 4), radius = √3

Answer 1

`Center(-1;4),`
`radius=√(3)`


`(x+1)^2+(y-4)^2=(√(3))^2`

`x^2+2x+1+y^2-8y+16=3`

`x^2+y^2+2x-8y+14=0`

:)

Answer 2

The equation of circle with Center (-1, 4), radius = √3 is
`(x+1)^2+(y-4)^2=3`

Explanation:

Given : Center (-1, 4), radius = √3

We have to find the equation of circle with Center (-1, 4), radius = √3.

The standard equation of circle with center (h,k) and radius r is given by

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

Given Center (-1, 4), radius = √3

Subsitute, h = -1 , k = 4 and r = √3

We have,

`(x-(-1))^2+(y-4)^2=(√(3))^2`

Simplify, we have,

`(x+1)^2+(y-4)^2=3`

Thus, The equation of circle with Center (-1, 4), radius = √3 is
`(x+1)^2+(y-4)^2=3`