Question

If \[{g^2} + {f^2} = c\] , then find the radius of the circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\] .
A. A circle of radius g.
B. A circle of radius of f
C. A circle of diameter \[\sqrt c \]
D. A circle of radius 0.

Answer 1

Hint: Write the general form of a circle, radius and centre of the circle. Then substitute the given result in the equation of the radius of the general form and simplify to obtain the required answer.

Formula used:
The general equation of a circle is,
\[{x^2} + {y^2} + 2gx + 2fy + c = 0\], where \[( - g, - f)\] is the centre and \[\sqrt {{g^2} + {f^2} - c} \] is the radius.

Complete step by step solution:
The general equation of a circle is,
\[{x^2} + {y^2} + 2gx + 2fy + c = 0\], where \[( - g, - f)\] is the centre and \[\sqrt {{g^2} + {f^2} - c} \] is the radius.
It is given that \[{g^2} + {f^2} = c\].
Therefore,
Substitute \[{g^2} + {f^2} = c\] in the equation of the radius \[\sqrt {{g^2} + {f^2} - c} \] to obtain the required answer.
Therefore, the radius is
 \[\sqrt {c - c} = 0\]
The correct option is D.

Additional information:
If the radius of a circle is zero, this means that the equation of the circle represents a point. A point has no dimension.

Note: There is another method to solve it. Substitute \[{g^2} + {f^2} = c\] in the equation \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]: \[{x^2} + {y^2} + 2gx + 2fy + {g^2} + {f^2} = 0\]. Simplify the equation: \[{\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} = 0\]. Now compare the equation with \[{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\], we get \[r = 0\]. Thus the radius of the circle is 0.