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12. [Ex 2K) Find the centre, C, and the radius, r, of the following equation of circle: x2 + y2 - 6x + 4y + 9 = 0

User Shenequa
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1 Answer

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An equation in the form:


(x-a)^2+(y-b)^2=r^2

is the standard form for the equation of a circle with center (a,b) and radius r. Here we have:


x^2+y^2-6x+4y+9=0

Then, group the x and y terms separately and "move" the constant to the right side of the equation:


x^2-6x+y^2+4y=-9

Complete the square:


x^2-6x+9+y^2+4y+4=-9+9+4

Factor:


(x-3)^2+(y+2)^2=4

Express the right side as a square:


(x-3)^2+(y-(-2))^2=2^2

Therefore:

The center is: (3, - 2), the radius is 2

Answer:


\begin{gathered} \text{Center: (3,-2)} \\ \text{Radius: 2} \end{gathered}

User NrNazifi
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