Question

Is the point $(3,5)$ inside or outside or on the circle with the equation $x^2+y^2=9$?

Answer 1

The center of the circle $x^2+y^2=9$ is $(0,0)$.

Let us write it as $x^2+y^2-9=0$ and take $f(x,y)=x^2+y^2-9=0$.

For the points lying on the circle, $f(x,y)=0$, for the points inside the circle, it has a sign as the sign obtained by putting the value of the coordinates of the center in the expression.

If it has opposite sign then lies outside the circle.

$f(0,0)=0+0-9=-9=$ negative

Given point is $(3,5)$

Hence, $f(3,5)$ is

$f(3,5)=3^2+5^2-9$

$f(3,5)=9+25-9$

$f(3,5)=25=$ positive

Therefore, the given point lies outside the circle.