Question

Find the centre and radius of the circle 3x2 +3y2-6x+4y-4

Answer 1

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Answer 2

Answer:

Centre of the circle  = (1,-$\frac{2}{3}$)

Radius of the circle = $\frac{5}{3}$

Step-by-step explanation:

Given,

The equation of a circle is 3x²+3y²-6x+4y-4

To find,

The center and radius of the circle

Recall the formula

The general equation of the circle is x² + y² + 2gx + 2fy + c = 0 -------(1)

with center =  (−g,−f)  and square of radius = a² = g² + f²− c.

Solution

Given equation is 3x²+3y²-6x+4y-4

Dividing through out by '3' we get

x²+y²-2x+$\frac{4}{3}$y-$\frac{4}{3}$  --------------------(2)

Comparing equations (1) and (2) we get

2g = -2 ⇒g = -1

2f = +$\frac{4}{3}$ ⇒f = +$\frac{2}{3}$

c = -$\frac{4}{3}$

Hence,

Centre of the circle =  (−g,−f) = (1,-$\frac{2}{3}$)

Square of radius of the circle = g² + f²− c.

=1² + ($\frac{2}{3}$)² - (-$\frac{4}{3}$ )

= 1 + $\frac{4}{9}$ +$\frac{4}{3}$

=$\frac{9+4+12}{9}$

=$\frac{25}{9}$

Radius = $\sqrt{\frac{25}{9} }$ = $\frac{5}{3}$

∴Centre of the circle  = (1,-$\frac{2}{3}$)

Radius of the circle = $\frac{5}{3}$

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