Question

Use the equation of the circle centered at the origin and

Answer 1

Answer:

Step-by-step explanation:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

If we are given x and y as 0 and 5 respectively, and we are also told that the center is (0, 0), our h and k are both 0. Filling in x, y, h, and k we can find the radius. So let's do that:

$(0-0)^2+(5-0)^2=r^2$ and

$0^2+5^2=r^2$ so

$r^2=25$  Our circle's equation is

$x^2+y^2=25$

Since the point in question, (4, 4), lies in the first quadrant, we will concentrate on that quadrant only. To fall within the circle, we can set up an inequality and test the point (4, 4). If it lies ON the circle then the equality would be true. Let's try that first:

$4^2+4^2=25$

Obviously, 16 + 16 does not equal 25, so that point (4, 4) does not lie ON the circle. In fact from that statement alone, we can determine that the point lies outside the circle because

$4^2+4^2>25$

If the inequality < were true then the point would lie inside.