Question

Why do we take the coordinates of the centre as (-g,-f) and why not (g,f) ?

Answer 1

Hint: The combination of points lying in the same plane having equal distance from a point called the centre is called a circle. The constant distance between centre and any point on the circle is called the radius of the circle. Compare the standard equation of the circle with its general equation to find out the coordinates of the centre of the circle.

Complete step-by-step answer:
We know the equation of a circle whose centre is at the origin is given by,
 $ {x^2} + {y^2} = {r^2} $
And that of a circle with centre at the point (h,k) is given by,
 $ {(x - h)^2} + {(y - k)^2} = {r^2} $
Expanding the above equation, we get –
 $ {x^2} + {h^2} - 2hx + {y^2} + {k^2} - 2ky = {r^2} $
But the general equation of a circle is –
 $ {x^2} + 2gx + {y^2} + 2fy + c = 0 $
On comparing the above two equations, we get –
 $
  c = {h^2} + {k^2} - {r^2} \\
  2gx = - 2hx \\
   \Rightarrow h = - g \\
  2fy = - 2ky \\
   \Rightarrow k = - f \;
  $
Thus, to satisfy the general equation of circle coordinates of the centre of the circle have to be (-g,-f) and cannot be (g,f).

Note: In the equation of circle when we replace h by –g and k by –f, we get,
 $ {x^2} + 2gx + {y^2} + 2fy + {g^2} + {f^2} - {r^2} = 0 $
The last three terms don’t involve x or y at all so they get replaced by a constant c.
The equation $ {x^2} + 2gx + {y^2} + 2fy + c = 0 $ is known as the general equation because it is quadratic in both x and y , there is no term in xy and the coefficient of both $ {x^2} $ and $ {y^2} $ is one.
The centre of the circle having the above equation is at $ ( - g, - f) $ and the radius of the circle is given as $ r = \sqrt {{g^2} + {f^2} - c} $ .