Question 1160508

an equation of the circle is:

{{{(x-h)^2+(y-k)^2=r^2}}} where {{{h}}} and {{{k}}} are coordinates of the center, and {{{r}}} is a radius

given:
the circle with center at the origin, means {{{h=0}}} and {{{k=0}}}

so far your equation is

{{{(x-0)^2+(y-0)^2=r^2}}}

{{{x^2+y^2=r^2}}}.....use given: circle  passes through the point P({{{5}}},{{{-3}}}) to calculate {{{r^2}}}

{{{5^2+(-3)^2=r^2}}}

{{{25+9=r^2}}}

{{{34=r^2}}}

and   your equation is

{{{x^2+y^2=34}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(5,-3,.12),locate(5,-3,P(5,-3)),
graph( 600, 600, -10, 10, -10, 10,-sqrt(34-x^2) ,sqrt(34-x^2))) }}}