The equation of circle in standard form is \((x-2)^{2}+(y+3)^{2} =(3\sqrt{2})^{2}\).
The center of the circle is at the point \((2,-3)\) and its radius \(3\sqrt{2}\) units.
Given equation of circle,
\(7x^2 + 7y^2 -28x + 42y -35 = 0.\)
We have to find the coordinate of center and its radius.
Now, first make the given equation in standard form, for this we have to divide the equation by 7 both sides,we get
\(x^{2} +y^{2}+-4x+6y-5=0\\\)
\(x^{2} -4x+y^{2} +6y=5\\\)
Now making the above equation in square form, we get
\(x^{2} -2\times x\times 2+2^{2}+y^{2}+2\times y\times3+3^{2} =5+4+9\\\)
\((x-2)^{2}+(y+3)^{2} =18\)
\((x-2)^{2}+(y+3)^{2} =(3\sqrt{2})^{2}\).........(1)
Hence the standard form of given equation is \((x-2)^{2}+(y+3)^{2} =(3\sqrt{2})^{2}\).
We know that, the general equation for a circle is \(( x - h )^2 + ( y - k )^2 = r^2\) where ( h, k ) is the center and r is the radius.
On comparing equation 1 with the general form of circle, we get coordinates if center\((2,-3)\) and radius is \(3\sqrt{2}\) .
Thus the center of the circle is at the point and its radius units.
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