A right triangle and a circle overlap on the coordinate plane. The three coordinates of the triangle are (0, 0), (r, 0), and (r, y), where r and y are positive constants, and the circle has a radius of r with a center at the origin. What is the length, in units, of the segment of the hypotenuse of the triangle that is outside the circle?
Answer (2)
Find the equation of the hypotenuse: y = r y x .
Find the equation of the circle: x 2 + y 2 = r 2 .
Find the intersection point of the hypotenuse and the circle: ( r 2 + y 2 r 2 , r 2 + y 2 ry ) .
Calculate the length of the segment of the hypotenuse outside the circle: r 2 + y 2 − r .
Explanation
Problem Analysis Let's analyze the problem. We have a right triangle with vertices at (0, 0), (r, 0), and (r, y), and a circle centered at the origin with radius r. We need to find the length of the part of the triangle's hypotenuse that lies outside the circle.
Equation of the Hypotenuse First, we need to find the equation of the hypotenuse. Since it passes through (0, 0) and (r, y), its equation is given by:
y = r y x
Equation of the Circle Next, we find the equation of the circle centered at the origin with radius r:
x 2 + y 2 = r 2
Finding the Intersection Points - x coordinate Now, we need to find the intersection points of the hypotenuse and the circle. Substitute the equation of the hypotenuse into the equation of the circle:
x 2 + ( r y x ) 2 = r 2 x 2 + r 2 y 2 x 2 = r 2 x 2 ( 1 + r 2 y 2 ) = r 2 x 2 = 1 + r 2 y 2 r 2 x 2 = r 2 + y 2 r 4 x = r 2 + y 2 r 2
Since we are looking for the intersection in the first quadrant (x > 0), we take the positive square root.
Finding the Intersection Points - y coordinate Now, we find the y-coordinate of the intersection point:
y = r y x = r y ⋅ r 2 + y 2 r 2 = r 2 + y 2 ry
So, the intersection point is ( r 2 + y 2 r 2 , r 2 + y 2 ry ) .
Length of the Hypotenuse Next, we calculate the length of the hypotenuse. The hypotenuse connects (0, 0) and (r, y), so its length is:
L h y p o t e n u se = ( r − 0 ) 2 + ( y − 0 ) 2 = r 2 + y 2
Length of the Hypotenuse Inside the Circle Now, we calculate the distance from the origin to the intersection point. This is the length of the segment of the hypotenuse inside the circle:
L in s i d e = ( r 2 + y 2 r 2 ) 2 + ( r 2 + y 2 ry ) 2 = r 2 + y 2 r 4 + r 2 + y 2 r 2 y 2 = r 2 + y 2 r 4 + r 2 y 2 = r 2 + y 2 r 2 ( r 2 + y 2 ) = r 2 = r
Length of the Hypotenuse Outside the Circle Finally, we subtract the length of the segment inside the circle from the total length of the hypotenuse to find the length of the segment outside the circle:
L o u t s i d e = L h y p o t e n u se − L in s i d e = r 2 + y 2 − r
Final Answer Therefore, the length of the segment of the hypotenuse that is outside the circle is r 2 + y 2 − r .
Examples
This problem combines geometry and algebra, which is useful in various real-world applications. For instance, when designing a circular garden with a straight path cutting across it, you might need to calculate how much of the path lies outside the garden. This involves finding the intersection of a line and a circle, similar to what we did in this problem. Another example is in optics, where you might need to calculate the length of a light ray that passes through a circular lens.
The length of the segment of the hypotenuse of the triangle that lies outside of the circle is r 2 + y 2 − r . This is calculated by first finding the total length of the hypotenuse and then subtracting the length that lies inside the circle. The calculations involve deriving the equations for both the hypotenuse and the circle and finding their intersection points.
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