## College Algebra 7th Edition

$x^2+y^2=65$
RECALL: The standard form of a circle whose center is at $(h, k)$ and radius $r$ is $(x--h)^2+(y-k)^2=r^2$. The center is at $(0, 0)$, then $h=0$ and $k=0$. Thus, the standard form of the given circle's equation of the circle is: $(x-0)^2+(y-0)^2=r^2 \\x^2+y^2=r^2$ The circle passes through the point $(4, 7)$. This means that the coordinates of this point satisfy the equation of the circle. Substitute the x and y values of the point into the equation to obtain: $x^2+y^2=r^2 \\4^2+7^2=r^2 \\16+49=r^2 \\65=r^2$ Thus, the standard form of the given circle's equation is: $x^2+y^2=65$