Question

Find the centre and radius of each of the circles whose equations
are given below 3x2+3y2+6x-12y-1=0

Answer 1

EXPLANATION.

Centre and Radis of the circle.

Equation = 3x² + 3y² + 6x - 12y - 1 = 0.

As we know that,

General equation of circle.

⇒ x² + y² + 2gx + 2fy + c = 0.

⇒ Centre of the circle = (-g,-f).

⇒ Centre of the circle = (-3,6).

⇒ Radius of circle = √g² + f² - c.

⇒ Radius of circle = √(-3)² + (6)² - (-1).

⇒ Radius of circle = √9 + 36 + 1.

⇒ Radius of circle = √46.

                                                                                                                   

MORE INFORMATION.

General equation of circle represents.

(1) = A real circle if : g² + f² - c > 0.

(2) = A point circle if : g² + f² - c = 0.

(3) = An imaginary circle if : g² + f² - c < 0.

Answer 2

Answer:

• Centre of Circle = (-3, 6)
• Radius of Circle = √46

Step-by-step explanation:

Given,

• Equation of given circle = 3x² + 3y² + 6x - 12y - 1 = 0

To Find,

• The centre and radius of circle.

Solution,

Equation of given circle;

3x² + 3y² + 6x - 12y - 1 = 0

Equation of Any circle in Form of;

→ x² + y² + 2gx + 2fy + c = 0

Comparing both Equations,

$: \implies {x}^{2} + {y}^{2} + 2gx + 2fy + c = 3 {x}^{2} + 3 {y}^{2} + 6x - 12y - 1 \\ \\ : \implies {x}^{2} + {y}^{2} + 2gx + 2fy + c = 3( {x}^{2}) + 3( {y}^{2}) + 2(3)x + 2( - 6)y + (- 1) \\ \\ : \implies \color{red}\boxed{g = 3} \\ \\ : \implies \color{red} \boxed{f = - 6} \\ \\ : \implies \color{red}\tt \boxed{c = - 1}$

Centre of Circle = (-g, -f)

• → Centre of Circle = (-(3), -(-6))
• → Centre of Circle = (-3, 6)

Radius of Circle = √(g² + f² - c)

• → Radius of Circle = √[(3)² + (-6)² - (-1)]
• → Radius of Circle = √[9 + 36 + 1]
• → Radius of Circle = √46

Required Answer,

• Centre of Circle = (-3, 6)
• Radius of Circle = √46