Find the intersecting point of yaxis for the xaxis of a circle

Question

I need to find the intersecting points of x and y axis in a circle for the given radius of 40. So I found the formula $(x-h)^2 + (y-k)^2 = r^2$, center of the circle is $(0,0)$. X-axis is from $-5$ to $5$ and y axis is from $-5$ to $5$. I want to find the intersecting point of y-axis for x-axis $2$. So I applied the formula like

$$(2-0)^2 + (y-k)^2 = (40)2$$ $$4 + y^2 = 1600$$ $$y^2 = 1600-4$$ $$y = 39.9499687109$$ but my y-axis is from -5 to +5 only. I have attached my work. How can I get the y axis value. What I'm doing wrong here? Please help me to solve this. Thank you.

Answer 1

As the radius is $40$ units, the intersection points are $(-40,0),(0,40),(40,0),(0,-40)$ and not $(0,\pm5)$ and $(\pm5,0)$. So at $x = 2$, $y \approx 39.949 $ is correct.

Answer 2

You correctly found the value of $y$ when $x=2$. (It's not the intersection with the $y$-axis, which would be at height $40$.)

That point does not show on your picture because the circle in your picture is much too small. Since the circle has radius $40$ it meets the $x$-axis at $40$ and $-40$, not as you have drawn it.

Answer 3

If a circle has radius $40$ then the distance from the origin along the $x$-axis to a point should be $40$ not $5$! Meaning that the intersections should be $(±40,0)$ and $(0,±40)$! I have attached a picture below]1

So when $x=2$, then $2^2+y^2=1600$ making $y=39.95$ as you said!