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Sagot :
Let's begin by listing out the given information:
[tex]\begin{gathered} 36x^2+36y^2-36x+48y=-16 \\ \text{dividing through by 36 to remove the coefficients of }x^2,y^2\text{, we have:} \\ \frac{36}{36}x^2-\frac{36}{36}x+\frac{36}{36}y^2+\frac{48}{36}y=-\frac{16}{36} \\ x^2-x+y^2+\frac{4}{3}y=-\frac{4}{9} \\ x^2-x+y^2+\frac{4}{3}y+\frac{4}{9}=0 \\ Comparing\text{ }this\text{ }with\text{ }the\text{ }general\text{ }equation\text{ of }circle\text{ }we\text{ have:} \\ x^2+y^2+2gx+2fy+c=0 \\ \Rightarrow2g=-1 \\ g=-\frac{1}{2} \\ 2f=\frac{2}{3} \\ \Rightarrow f=\frac{1}{3} \\ c=\frac{4}{9} \\ \Rightarrow c=\frac{4}{9} \end{gathered}[/tex]Thus the center of the circle is:
[tex]\begin{gathered} (h,k)=(-g,-f) \\ \Rightarrow(--\frac{1}{2},-\frac{1}{3}) \\ \Rightarrow(\frac{1}{2},-\frac{1}{3}) \\ \\ r=\sqrt{g^{2}+f^{2}-c} \\ r=\sqrt[]{(\frac{1}{2})^2+(-\frac{1}{3})^2-\frac{4}{9}} \\ r=\sqrt[]{\frac{1}{4}+\frac{1}{9}-\frac{4}{9}}=\sqrt[]{-\frac{1}{12}} \\ r=\sqrt[]{-\frac{1}{12}} \end{gathered}[/tex]
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