Question

write the radius of the general equation of the circle x²+y²+2gx+2fy+c=0 is​

Answer 1

Concept: Equation of circle having centre (h,k) and radius r is (x - h)2 + (y - k)2 = 0

Calculation:

x² + y²- 2hx - 2hy + h2 + k2 = r2

x² + y2²- 2hx - 2hy + h² + k²- r² = 0

Comparing the above equation with x² + y² + 2gx + 2fy + c = 0, we get,

h = -g, k = -f and h² + k²- r²= c

Now, (x² + 2gx + g²) + (y² + 2fy + f²) = g²+ f²- c

(x + g)2 + (y + f)2 =$\sqrt{ {g}^{2} + {f}^{2} - c }$

${x - (-g)}2 + {y - (-f)}2 =\sqrt{ {g}^{2 } + {f}^{2} - c }$

The equation x² + y² + 2gx + 2fy + c = 0 always represents a circle whose centre is (-g, -f), that is ($- \frac{1}{2}$coefficient of x,$- \frac{1}{2}$coefficient of y) and radius is

$\sqrt{ {g}^{2} + {f}^{2} - c }$

If g² + f² - c > 0 then the radius of the circle becomes zero (degenerate circle). In this case, the circle reduces to the point (-g, -f). Such a circle is known as a point circle. In other words, the equation x² + y²+ 2gx + 2fy + c = 0 represents a point circle.

If g² + f² - c > 0, then the radius of the circle is real and hence the equation x² + y² + 2gx + 2fy + c = 0 represents a real circle.

If g² + f²- c < 0, the radius of the circle$\sqrt{ {g}^{2} + {f}^{2} } -$becomes imaginary but the circle is real. Such a circle is called an imaginary circle. In other words, equation x² + y² + 2gx + 2fy + c = 0 does not represent any real circle as it is not possible to draw such a circle.