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]
