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Earlier discussion
2 older comments kept for context.
Victor Titus
The diameter of the circle given by the equation \(2x^{2} + 2y^{2} - 50 = 0\) is 10 units.
Here's how we find it:
The general form of the equation of a circle in Cartesian coordinates is \(x^{2} + y^{2} = r^{2}\), where \(r\) is the radius of the circle.
In the given equation, we can divide through by 2 to get \(x^{2} + y^{2} = 25\), which shows that the radius of the circle is \(\sqrt{25} = 5\) units.
The diameter of a circle is twice its radius, so the diameter of this circle is \(2 \times 5 = 10\) units.
Oluwaseun Adeyemi
The general equation of a circle is given as \(x^2+y^2+2gx+2fy+c = 0\)
Where \(a = -g, b = -f\) and \(c = (a^{2} + b^{2}) - r^{2}\).
By comparing the standard equation of a circle \([x2+y2+2\mathit{gx}+2\mathit{fy}+c=0]\) and the given equation of the circle \((2x^2+2y^2-50=0)\) the values of g, f and c can be obtained. But before the comparison the two equations must be in the same form.
\(2^2+2y^2-50 = 0\)
Divide through the equation by 2, \(x^2+y^2-25=0\)
By comparison:
\(g=0,f=0,c=-25\), But \(a =-g=0, b= -f =0\)
\(c=a^2+b^2-r^2\), \(-25=a^2+b^2-r^2\)
\(-25=0^2+0^2-r^2\), \(-25=r^2,r^2=25\)
\(r^2=\sqrt{25}=5\)
Diameter(d) \(= 2r = 2 \times 5=10\) units