Question

Find the center and radius of the circle

Answer 1

Centre:

• The general formula for the circumference is:

$\boxed{ \boxed{{x^{2} \: + \: y ^{2} \: + \: Dx \: + \: Ey \: + F \: = \: 0}}}$

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To find the center, write this formula:

$\boxed{ \boxed{C( \frac{-D}{2} , \frac{-E}{2} )}}$

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We know that...

$D \: = \: -6 \\ E \: = \: 4 \\ F \: = \: 4$

We use the equation of the center with the values already obtained:

$C( \frac{- - 6}{2} , \frac{-4}{2} )$

$C( \frac{6}{2} , \frac{-4}{2} )$

$\huge\boxed{ \bold{C( 3, - 2 )}}$

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Ratio:

Now we use the equation of the radius for a circumference, which is:

$\boxed{ \boxed{{r \: = \: \frac{1}{2} \sqrt{D ^{2} \: + \: E ^{2} \: - \: 4F } }}}$

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Now we use the equation of the radius for a circumference with the values already obtained.

$r \: = \: \frac{1}{2} \sqrt{ { - 6}^{2} \: + \: {4}^{2} \: - \: 4 \: \times \: 4 }$

• I am going to use complex numbers because the square root of a negative number does not exist in the set of real numbers.

$r \: = \: \frac{1}{2} \sqrt{ { - 6}^{2} \: - \: {4}^{2} \: - + \: 4 \: \times \: 4 i}$

$r \: = \: \frac{1}{2} \sqrt{ { 6}^{2} \: - \: {4}^{2} \: + \: 16i}$

$r \: = \: \frac{ \sqrt{ {6}^{2} \: - \: {4}^{2} \: + \: 16i} }{2}$

$r \: = \: \frac{ \sqrt{36 \: - \: 16 \: + \: 16i} }{2}$

$r \: = \: \frac{ \sqrt{36i} }{2}$

$r \: = \: \frac{6i}{2}$

$\huge \boxed{ \bold{r \: = \: 3i}}$

MissSpanish

Answer 2

AnswEr :

Provided Equation

• x² + y² - 6x + 4y + 4= 0

As we know the Standard Equation of a Circle is

$\bf{(x - a)} {}^{2} + (y - b) {}^{2} = {r}^{2}$

where,

r radius of circle.

(a,b) centre .

$\implies \sf \: {x}^{2} + {y}^{2} - 6x + 4y + 4 = 0$

This equation can be further written as

$\implies \sf \: {x}^{2} - 6x + {y}^{2} + 4y = - 4$

Now completing the square ( by adding 4 & 9 on both side ) .

$\implies \sf \: {x}^{2} - 6x + 9 + {y}^{2} + 4y + 4 = - 4 + 9 + 4$

again this equation can be further written as

$\implies \sf \: (x - 3) {}^{2} + (y + 2) {}^{2} = {3}^{2}$

Now comparing this equation with standard equation of circle ( mentioned above) and we will get

• Centre = ( a,b) = (3, -2 )
• Radius = r = 3

Therefore,

• Centre of circle is (3,-2) and radius is 3