A Circle And A Line

Geometrically, $x^2+y^2=9$ is a circle of radii $3$ and center on $(0,0)$ so the circle doesn't go beyond $y=3$ and $x=3\therefore y=4$ doesn't meet with $x^2+y^2=9$

What else to say...

When $y=4$ , $x^2=-7$ which means $x$ is a complex number, hence $y=4$ does not intersect the given graph at all.

Way 1 to solve : Because x^2 >= 0 so that y^2 <=9 so that -3 <= y <= 3, its graph can not meet the line y =4

Way 2 to solve : with y =4 , x^2 = -7 <0 --> the point that the graph meet the line doesn't exist.

Obviously, the value can never go beyond the square of 4.

Just solve for $x , y$ from the two equations $y = 4 \ \ \Rightarrow (1) \ \ , \ \ \ \ x^2 + y^2 = 9 \ \ \Rightarrow (2)$ and then the outcome of the ordered pairs $(x,y)$ will be the points of intersection , and here is how :

From $(1)$ we see that always $y = 4$

And by substituting $y$ from $(1)$ into $(2)$ , we get $x^2 + y^2 = x^2 + 4^2 = 9$

$\therefore x^2 = 9 - 4^2 = -7$ $\therefore x = \pm \sqrt{-7} = \pm \sqrt{7} i$ Which are apparently imaginary numbers .

But the graph requires a real values for $x,y$ in order to get real points $(x,y)$ .

Which means that there's no real ordered pairs $(x,y)$ satisfying the two equations , and hence they have no meeting point on graph . i.e. $\boxed{0}$

That eqn is a circle of radius 3 units.. And a line parallel to x axis at a distance of 4 units from origin can't pass through the circle.

Y= 4 is the y intercept parallel to x axis. Will never touch the circle of radius 3 units.

The graph of the 2nd equation is a a circle with radius 3, and the graph of the first equation is a line with y-intercept at (0,4). The line is beyond the circumference of the circle