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Q4 O center (-1,3), radius = 1 What is the center and radius of the circle, 2 2 r + y + 2x - 6y +9=0 - 9 center (1,-3), radius = 1 center (-1,3), radius = 9 center (1,-3), radius = 3

Sagot :

Step 1: Write out the formula

The equation of a circle given by

[tex](x-a)^2+(y-b)^2=r^2[/tex][tex]\begin{gathered} \text{where} \\ (a,b)\text{ is the center of the circle} \\ r\text{ is the radius of the circle} \end{gathered}[/tex]

Step 2: Write out the given equation and rewrite it in the form shown above

[tex]x^2+y^2+2x-6y+9=0[/tex][tex]\begin{gathered} x^2+2x+y^2-6y+9=0 \\ \text{ By completing the square, we have} \\ (x+1)^2-(+1)^2+(y-3)^2-(-3)^2+9=0 \\ \end{gathered}[/tex][tex]\begin{gathered} (x+1)^2+(y-3)^2-1-9+9=0 \\ (x+1)^2+(y-3)^2=1=1^2 \end{gathered}[/tex]

By comparing the equation with the formula above, we have

[tex]a=-1,b=3,r=1[/tex]

Therefore,

center (-1,3), radius = 1

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