Question

whats the answer to this

Answer 1

Answer:

2

Step-by-step explanation:

equation of a circle =>

x^2 + y^2 + 2gx + 2hy + c = 0

comparing the given equation,

2g = 4

g = 2

2h = -2

h = -1

radius of a circle is given as

r = (√( g^2 + f^2 - c))

= (√( 2^2 + 1^2 - 1))

= (√( 4 + 1 - 1))

= (√(4))

= 2

hence, the radius of the circle is 2

hope it helps you

Answer 2

Answer:

2

Step-by-step explanation:

We know,

     Equation of a circle are written in the form of

     √{ ( x - x₁ )^2 + ( y - y₁ )^2 } = radius

Here,

⇒ x^2 + y^2 + 4x - 2y = - 1

⇒ x^2 + 4x + y^2 - 2y = - 1

     Adding 4 to both sides :

⇒ x^2 + 4x + 4 + y^2 - 2y = - 1 + 4

     Adding 1 to both sides;

⇒ x^2 + 4x + 4 + y^2 - 2y + 1 = - 1 + 1 + 4

⇒ ( x^2 + 4x + 4 ) + ( y^2 - 2y + 1 ) = 4

      Using a^2 + b^2 + 2ab = ( a + b )^2

                 a^2 + b^2 - 2ab = ( a - b )^2

⇒ ( x + 2 )^2 + ( y - 1 )^2 = 4

⇒ { x - ( - 2  ) }^2 + ( y - 1 )^2 = 2^2

⇒ √[ { x - ( - 2 ) }^2 + ( y - 1 )^2 ] = 2

        Now, this equation is in the form as given above, so radius of the that circle must have been represented by the RHS of this eq.

       Radius = 2