Write the general equation for the circle that passes through the points: (0, 0) (6, 0)(0, - 8)You must include the appropriate sign (+ or -) in your answer. Do not use spaces in your answer.x^2 + y^2 ____ x ____ y = 0Please Help!

Answer:[tex]x^{2} + y^{2} -6x+8y= 0[/tex]Step-by-step explanation:The general equation of the circle is of the form:[tex]x^{2} + y^{2} +2gx+2fy + c= 0[/tex]The circle passes through the point (0,0), this means replacing x = 0 and y = 0 must satisfy the equation Using these values, we get:[tex]0^{2} +0^{2}+2g(0)+2f(0)+c=0\\c=0[/tex]This means value of c is zero for the given circle. So, now the equation of the circle is:[tex]x^{2} + y^{2} +2gx+2fy= 0[/tex]Now using the point, (6, 0) in this equation, we get:[tex]6^{2}+ 0^{2}+2g(6)+2f(0)=0\\36+12g=0\\36=-12g\\g=-3[/tex]Hence the value of g is -3, using the value of g and c in our equation, the equation becomes:[tex]x^{2} + y^{2} -6x+2fy= 0[/tex]Now using the 3rd point (0, -8) in this equation to find the value of f:[tex]0^{2}+ (-8)^{2}-6(0)+2f(-8)=0\\64-16f=0\\64=16f\\f=4[/tex]Using the value of g, f and c, the final equation of the circle is:[tex]x^{2} + y^{2} -6x+8y= 0[/tex]