Given:
The general equation of a circle is,
[tex]7x^2+7y^2-28x+42y-35=0\text{ ------(1)}[/tex]
The standard equation of a circle is given by,
[tex](x-h)^2+(y-k)^2=r^2\text{ ------(2)}[/tex]
Here, (h, k) is coordinates of the center of the circle and r is the radius of the circle.
Equation (1) can be converted to the standard form by first dividing equation (1) by 7.
[tex]x^2+y^2-4x+6y-5=0\text{ }[/tex]
Grouping the x's and y's in the above equation,
[tex]x^2-4x+y^2+6y=5[/tex]
Complete the square of variables x and y by adding constants that create prefect square trinomials.
[tex]\begin{gathered} x^2-4x+4+y^2+6y+9=5+4+9 \\ x^2-2\times x\times2+2^2+y^2+2\times y\times3+3^2=18 \end{gathered}[/tex]
Now, factor the perfect square trinomials in each variable.
[tex]\begin{gathered} (x-2_{})^2+(y+3)^2=(\sqrt[]{18})^2 \\ (x-2_{})^2+(y+3)^2=(3\sqrt[]{2})^2 \end{gathered}[/tex]
So, the equation of the circle in standard form is,
[tex](x-2_{})^2+(y+3)^2=(3\sqrt[]{2})^2[/tex]
Comparing the above equation with the standard form of the equation of a circle given in equation (2), we get
[tex]h=2,\text{ k=-3 and r=3}\sqrt[]{2}[/tex]
Therefore, the center of the circle is at the point (2, -3) .
The radius of the circle is,
[tex]3\sqrt[]{2}[/tex]
