The equation of circle in standard form is [tex](x-2)^{2}+(y+3)^{2} =(3\sqrt{2})^{2}[/tex].
The center of the circle is at the point [tex](2,-3)[/tex] and its radius [tex]3\sqrt{2}[/tex] units.
Given equation of circle,
[tex]7x^2 + 7y^2 -28x + 42y -35 = 0.[/tex]
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
[tex]x^{2} +y^{2}+-4x+6y-5=0\\[/tex]
[tex]x^{2} -4x+y^{2} +6y=5\\[/tex]
Now making the above equation in square form, we get
[tex]x^{2} -2\times x\times 2+2^{2}+y^{2}+2\times y\times3+3^{2} =5+4+9\\[/tex]
[tex](x-2)^{2}+(y+3)^{2} =18[/tex]
[tex](x-2)^{2}+(y+3)^{2} =(3\sqrt{2})^{2}[/tex].........(1)
Hence the standard form of given equation is [tex](x-2)^{2}+(y+3)^{2} =(3\sqrt{2})^{2}[/tex].
We know that, the general equation for a circle is [tex]( x - h )^2 + ( y - k )^2 = r^2[/tex] 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[tex](2,-3)[/tex] and radius is [tex]3\sqrt{2}[/tex] .
Thus the center of the circle is at the point and its radius units.
For more details on general equation of circle follow the link:
https://brainly.com/question/10165274
