Question

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

Answer 1

eqn will be (x-(-2))^2+(y-1)^2=9
x^2+4+2x+y^2+1-2y=9
x^2+y^2+2x-2y-4=0

Answer 2

Answer:

The equation for the circle will be $x^2+y^2+4x-2y-4=0$

To find:

Equation = ?

Solution:

The circle is centred at the point (-2, 1) and has a radius of length 3.

Equation will be as under:

Equation for the circle will be,

$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }$

Given that, the radius is 3 and the centre is (-2, 1).

Thus, a = -2 and b = 1. The radius r = 3

The equation becomes,

$\begin{array} { l } { ( x - ( - 2 ) ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 9 } \\\\ { ( x + 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 9 } \\\\ { x ^ { 2 } + 4 + 4 x + y ^ { 2 } + 1 - 2 y = 9 } \\\\ { x ^ { 2 } + y ^ { 2 } + 4 x - 2 y + 5 = 9 } \\\\ { x ^ { 2 } + y ^ { 2 } + 4 x - 2 y + 5 - 9 = 0 } \\\\ { x ^ { 2 } + y ^ { 2 } + 4 x - 2 y - 4 = 0 } \end{array}$

Thus, the equation for the circle will be $x^2+y^2+4x-2y-4=0$