Answer:
\(x^2+y^2-41=0\)
Step-by-step explanation:
Center of the circle is (0,0)
the point on the circle is (4,5)
Center radius form of the circle is
\((x-h)^2 +(y-k)^2= r^2\)
where (h,k) is the center and 'r' is the radius of the circle
WE know center is (0,0) . Lets find out the radius using the given point
\((x-0)^2 +(y-0)^2= r^2\)
\(x^2+y^2= r^2\)
Plug in the given point (4,5)
\(4^2+5^2= r^2\)
\(16+25= r^2\)
\(41=r^2\)
Equation becomes
\(x^2+y^2=41\)
General form of equation is
\(x^2+y^2+gx+fx+c=0\)
\(x^2+y^2=41\)
\(x^2+y^2-41=0\)
