Question

Write the formula to find the centre and radius of the circle
x² + y² + 2gx + 2fy + c = 0 ?? ​

Answer 1

Answer:

Step-by-step explanation:

Conversely, a quadratic equation in x and y of the form x2 + y2

+ 2gx + 2fy + c = 0 always represents the equation of a circle.

We know that the equation of the circle having centre at (h, k) and radius = r units is

(x - h)2

+ (y - k)2 = r2

⇒ x2

+ y2 - 2hx - 2hy + h2 + k2 = r2

⇒ x2

+ y2 - 2hx - 2hy + h2 + k2 - r2

= 0

Compare the above equation x2

+ y2 - 2hx - 2hy + h2 + k2 - r2 = 0 with x2 + y2 + 2gx + 2fy + c = 0 we get, h = -g, k = -f and h2 + k2 - r2

= c

Therefore the equation of any circle can be expressed in the form x2

+ y2

+ 2gx + 2fy + c = 0.

Again, x2

+ y2

+ 2gx + 2fy + c = 0

⇒ (x2

+ 2gx + g2) + (y2 + 2fy + f2) = g2 + f2

- c

⇒ (x + g)2

+ (y + f)2 = (g2+f2−c‾‾‾‾‾‾‾‾‾‾‾√)2

⇒ {x - (-g) }2

+ {y - (-f) }2 = (g2+f2−c‾‾‾‾‾‾‾‾‾‾‾√)2

This is of the form (x - h)2

+ (y - k)2 = r2 which represents a circle having centre at (- g, -f) and radius g2+f2−c‾‾‾‾‾‾‾‾‾‾‾√

.

Hence the given equation x2

+ y2 + 2gx + 2fy + c = 0 represents a circle whose centre is (-g, -f) i.e, (-12 coefficient of x, -12 coefficient of y) and radius = g2+f2−c‾‾‾‾‾‾‾‾‾‾‾√ = (12coefficient of x)2+(12coefficient of y)2−constant term‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√