Triangle Angle & Circle Equation: A Math Guide

Hey math enthusiasts! Let's dive into a fun geometry problem. We're given three points: $P(-5,6)$, $Q(-3,8)$, and $R(3,2)$. Our mission, should we choose to accept it, is twofold: first, prove that the angle $PQR$ is a right angle, and second, find the equation of the circle that gracefully passes through all three points. Buckle up, because we're about to embark on a journey through slopes, right angles, and circles!

Proving Angle $PQR$ is a Right Angle

Alright, guys, let's tackle the first part of our challenge: demonstrating that angle $PQR$ is a right angle. The key here is to use the concept of slopes. Remember, the slope of a line tells us how steep it is. If two lines are perpendicular (and thus form a right angle), their slopes have a special relationship: they are negative reciprocals of each other. This means if one slope is $m$, the other slope is $-1/m$.

To find the slopes, we'll use the slope formula: $m = (y_2 - y_1) / (x_2 - x_1)$. Let's start by finding the slope of line segment $PQ$. We'll use the coordinates of $P(-5,6)$ and $Q(-3,8)$. Plugging these values into the formula, we get:

$m_{PQ} = (8 - 6) / (-3 - (-5)) = 2 / 2 = 1$

So, the slope of $PQ$ is 1. Now, let's find the slope of line segment $QR$. Using the coordinates of $Q(-3,8)$ and $R(3,2)$, we get:

$m_{QR} = (2 - 8) / (3 - (-3)) = -6 / 6 = -1$

Okay, check this out! The slope of $PQ$ is 1, and the slope of $QR$ is -1. Notice anything? These slopes are negative reciprocals of each other (1 and -1). The product of the slopes is -1. This tells us that the lines $PQ$ and $QR$ are perpendicular, which means they meet at a right angle at point $Q$. Therefore, angle $PQR$ is indeed a right angle! We've successfully completed the first part of our mission.

To drive the point home, think of it this way: a slope of 1 means the line goes up 1 unit for every 1 unit it moves to the right. A slope of -1 means the line goes down 1 unit for every 1 unit it moves to the right. When these two lines meet, they form a perfect right angle, just like the corner of a square. We used the slope formula to calculate the steepness of each line segment, and then we confirmed that the product of the slopes of the lines forming the angle $PQR$ is -1. This confirmed that angle $PQR$ is indeed a right angle. We've proven that the angle $PQR$ is a right angle by leveraging the relationship between slopes of perpendicular lines. This fundamental understanding of slopes is critical in geometry and is useful for many applications.

Let's summarize the key steps. First, we calculated the slope of $PQ$ using the coordinates of $P$ and $Q$. Second, we calculated the slope of $QR$ using the coordinates of $Q$ and $R$. Third, we observed that the slopes were negative reciprocals of each other, confirming that the lines were perpendicular, and the angle was a right angle. The slope formula is your best friend in these situations.

Now we know that $\angle PQR = 90^\circ$. The points P, Q and R forms a right angled triangle.

Finding the Equation of the Circle

Alright, now for the grand finale: finding the equation of the circle that passes through points $P$, $Q$, and $R$. Since we know that angle $PQR$ is a right angle, we're in luck! In a right-angled triangle, the circumcenter (the center of the circle that passes through all three vertices) is the midpoint of the hypotenuse. The hypotenuse in this case is $PR$.

So, to find the center of our circle, we need to find the midpoint of $PR$. The midpoint formula is: $((x_1 + x_2)/2, (y_1 + y_2)/2)$. Using the coordinates of $P(-5,6)$ and $R(3,2)$, we get:

Center = $((-5 + 3)/2, (6 + 2)/2) = (-2/2, 8/2) = (-1, 4)$

So, the center of our circle is at the point $(-1, 4)$. Now, we need to find the radius of the circle. The radius is the distance from the center to any of the points on the circle. Let's find the distance from the center $(-1, 4)$ to point $Q(-3,8)$. We'll use the distance formula: $d = \sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)$.

Radius = $\sqrt((-3 - (-1))^2 + (8 - 4)^2) = \sqrt((-2)^2 + 4^2) = \sqrt(4 + 16) = \sqrt(20)$

So, the radius of our circle is $\sqrt{20}$. Now that we have the center $(-1, 4)$ and the radius $\sqrt{20}$, we can write the equation of the circle. The standard equation of a circle is: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

Plugging in our values, we get: $(x - (-1))^2 + (y - 4)^2 = (\sqrt{20})^2$. This simplifies to: $(x + 1)^2 + (y - 4)^2 = 20$

And there you have it, guys! The equation of the circle that passes through points $P$, $Q$, and $R$ is $(x + 1)^2 + (y - 4)^2 = 20$. We successfully determined the equation of a circle that encompasses the points P, Q, and R. We've used the property that the hypotenuse of the right-angled triangle is the diameter of the circumcircle. The center of the circle is the midpoint of the hypotenuse. We found the center using the midpoint formula, and then the radius by calculating the distance between the center and a point on the circle, using the distance formula. The final equation of the circle is derived by substituting the coordinates of the center and the value of radius in the general equation of a circle. We've conquered both parts of the problem! We used the distance formula to find the radius of the circle. This is all about applying the distance formula and the midpoint formula.

Let's summarise the process. First, we identified that the center of the circle is the midpoint of the hypotenuse PR. Second, we used the midpoint formula to determine the coordinates of the center. Third, we found the radius by calculating the distance from the center to one of the points on the circle, using the distance formula. Finally, we substituted the center's coordinates and the radius in the general equation of a circle.

Conclusion

We did it! We successfully showed that angle $PQR$ is a right angle by analyzing the slopes of the line segments. We used the slope formula to calculate the slopes, and determined that the lines were perpendicular because their slopes were negative reciprocals of each other. Furthermore, we found the equation of the circle passing through points $P$, $Q$, and $R$. Because we knew it was a right-angled triangle, we utilized the midpoint of the hypotenuse to find the circle's center, then used the distance formula to find the radius and create the circle's equation. This approach combines slope analysis and circle geometry concepts. This problem showcases how different areas of mathematics are interconnected, allowing us to solve complex problems by applying various formulas and theorems. Keep practicing, and you'll be a geometry guru in no time!

Remember, understanding the relationships between slopes, right angles, and circles is key. Keep these formulas handy, practice with different examples, and you'll be well on your way to mastering these concepts. Congrats on sticking with it and tackling this problem. Remember that in mathematics, practice makes perfect. Keep exploring, keep learning, and don't be afraid to challenge yourselves. You've got this!