How to find the intercepts of a circle?
Understand the Problem
The question is asking how to determine the intercepts of a circle, which typically refers to the points where the circle intersects the axes in a Cartesian coordinate system. To solve this, we need the equation of the circle, and then we can set the equation equal to zero to find the x-intercepts and y-intercepts.
Answer
The x-intercepts are \( x = h \pm \sqrt{r^2 - k^2} \) and the y-intercepts are \( y = k \pm \sqrt{r^2 - h^2} \).
Answer for screen readers
The x-intercepts of the circle are ( x = h \pm \sqrt{r^2 - k^2} ) and the y-intercepts are ( y = k \pm \sqrt{r^2 - h^2} ).
Steps to Solve
- Identify the equation of the circle
The standard form of the equation of a circle is given by:
$$ (x - h)^2 + (y - k)^2 = r^2 $$
where ( (h, k) ) is the center of the circle and ( r ) is the radius.
- Find the x-intercepts
To find the x-intercepts, we set ( y = 0 ) in the equation of the circle. This changes the equation to:
$$ (x - h)^2 + (0 - k)^2 = r^2 $$
which simplifies to:
$$ (x - h)^2 + k^2 = r^2 $$
Next, solve for ( x ):
$$ (x - h)^2 = r^2 - k^2 $$
By taking the square root of both sides, we get:
$$ x - h = \pm \sqrt{r^2 - k^2} $$
Thus, the x-intercepts are:
$$ x = h \pm \sqrt{r^2 - k^2} $$
- Find the y-intercepts
To find the y-intercepts, we set ( x = 0 ) in the circle's equation. This gives us:
$$ (0 - h)^2 + (y - k)^2 = r^2 $$
which simplifies to:
$$ h^2 + (y - k)^2 = r^2 $$
Now, solve for ( y ):
$$ (y - k)^2 = r^2 - h^2 $$
Taking the square root of both sides yields:
$$ y - k = \pm \sqrt{r^2 - h^2} $$
Thus, the y-intercepts are:
$$ y = k \pm \sqrt{r^2 - h^2} $$
The x-intercepts of the circle are ( x = h \pm \sqrt{r^2 - k^2} ) and the y-intercepts are ( y = k \pm \sqrt{r^2 - h^2} ).
More Information
The intercepts tell us where the circle intersects the x-axis and y-axis. This information can be useful in graphing the circle or understanding its position relative to the axes. Finding intercepts is a common step in analyzing functions and curves.
Tips
- Forgetting to set ( y = 0 ) for x-intercepts and ( x = 0 ) for y-intercepts.
- Not applying the square root properly and missing the ( \pm ) sign when solving for intercepts.
- Assuming there are always real intercepts; sometimes, the square root results in a negative value, indicating that there are no real intercepts.
