Answer
Explanation
Given equation:
[tex]7x^2+7y^2-28x+42y-35=0[/tex]Step-by-step explanation:
The equation of a circle in standard form is given by:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ whereh,(h,k)\text{ }are\text{ }the\text{ }coordinates\text{ }of\text{ }the\text{ }centre,\text{ }and\text{ } \\ r\text{ }is\text{ }the\text{ }radius \end{gathered}[/tex]Divide the give equation through by 7:
[tex]\begin{gathered} \frac{7x^2}{7}+\frac{7y^2}{7}-\frac{28x}{7}+\frac{42y}{7}-\frac{35}{7}=\frac{0}{7} \\ \\ x^2+y^2-4x+6y-5=0 \end{gathered}[/tex]TCombine the like terms:
[tex]x^2-4x+y^2+6y=5[/tex]Using perfect square on each variable:
[tex]\begin{gathered} (x^2-4x+4)+(y^2+6y+9)=5+4+9 \\ \\ (x-2)^2+(y+3)^2=18 \\ \\ \therefore(h,k)=(2,-3) \\ \\ r^2=18 \\ \Rightarrow r=\sqrt{18}=3\sqrt{2} \end{gathered}[/tex]Therefore, the equation of this circle is
