Question

find the equation of circle, whose center is (-1,2) and radius is 3​

Answer 1

Answer:

x^2  + y^2 + 2x - 4y = 4

Step-by-step explanation:

equation of the circle is (x-h)^2  + (y-k)^2 = r^2

given h = -1 , k = 2 , r = 3

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

x^2 + 2x + 1 + y^2 - 4y +4  = 9

x^2  + y^2 + 2x - 4y + 5 = 9

x^2  + y^2 + 2x - 4y = 4   --> Answer

Answer 2

Answer:

Correct option is

B

x

2

+y

2

−4x+2y−4=0

Equation of circle with center (h,k) and radius r is written as

(x−h)

2

+(y−k)

2

=r

2

then,

Equation of circle with center (2,−1) and radius 3 is written as

⇒(x−2)

2

+(y+1)

2

=3

2

⇒x

2

+4−4x+y

2

+1+2y=9

∴x

2

+y

2

−4x+2y−4=0