Question

Prove that
x² + y² + 2gx + 2fy + c = 0

Answer 1

This equation may be written Hence (1) represents a circle whose centre is the point (- g, - f), and whose radius is If g2 + f2 > c, the radius of this circle is real. If g2 + f2 = c, the radius vanishes, i. e. the circle becomes a point coinciding with the point (- g, - f). Such a circle is called a point-circle. If g2 + f2 < c, the radius of the circle is imaginary.

OR:-

Theorem:- Prove that the equation

always represent a circle whose centre is

and radius

Answer 2

$\large\bf{\underline{\underline{Question:−}}}$

Prove that $\sf\color{red}{x² + y² + 2gx + 2fy + C = 0}$ represents a circle.Also, find its centre and radius.

$\large\bf{\underline{\underline{Solution:-}}}$

$\bold{We\: have,}$

$\sf{x² + y² + 2gx + 2fy + C = 0}$

$\sf{x² + y² + 2gx + 2fy = - C}$

$\sf{(x² + 2gx) + (y² + 2fy) = -C}$

$\sf{(x² + 2.x.g + g²) + (y² + 2.y.f + f²) = g² + f² - C}$

$\sf{ (x + g)² + (y + f)² = g² + f² - C}$

${x - (-g)}² + {y - (-f)}² = \sqrt{g² + f² - C}^{2}$

which is the form of$\sf\color{red}{(x - h)² + (y - k)² = r²}$

This shows that $\sf\color{red}{x² + y² + 2gx + 2fy = 0}$ represents a circle.

This type of equation of a circle is called general equation of the circle.

$\tt\color{orchid}{Hence,}$

${the \: centre \: and \: radius \: of \: the \: circle \: are (-g, -f) and \sqrt{g² + f² -C}}$

Thankyou :)