Problems from Equation of the circumference II: general equation

Considering the circumference $x^{2} + y^{2} + 2 x - 4 y + 4 = 0$ , find its radius and its center.

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Development:

In this case we have $A = - 2$ , $B = 4$ and $C = - 4$ . Therefore, the center will be $(\frac{- A}{2}, \frac{- B}{2}) = (\frac{2}{2}, \frac{- 4}{2}) = (1, - 2)$ and the radius will be $r = \sqrt{(\frac{A}{2})^{2} + (\frac{B}{2})^{2} - C} = \sqrt{(\frac{- 2}{2})^{2} + (\frac{4}{2})^{2} + 4} =$ $= \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Solution:

Center $(1, - 2)$ and radius $3$ .

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Find the equation of the circumference with center $(3, 4)$ and radius $2$ .

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Development:

$(x - 3)^{2} + (y - 4)^{2} = 2^{2} \Rightarrow x^{2} + y^{2} - 6 x - 8 y + 9 + 16 - 4 = 0 \Rightarrow$ $\Rightarrow x^{2} + y^{2} - 6 x - 8 y + 21 = 0$

Solution:

$(x - 3)^{2} + (y - 4)^{2} = 4$ or also $x^{2} + y^{2} - 6 x - 8 y + 21 = 0$

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