Finding 'a' & Locus Of Points: Mastering Circle Equations & Geometric Principles
Hey guys! Let's dive into some cool math problems that involve circles, points, and a little bit of geometry. We're going to figure out how to find the value of a variable when a point sits on a circle and then explore something called a 'locus,' which is like the path a point takes based on some rules. It's all pretty fun, I promise!
Determining the Value of 'a' When a Point Lies on a Circle
Okay, so first up, we have to determine the value of 'a' if a point is chilling on a circle. Remember, a circle equation is usually written as x² + y² = r², where 'r' is the radius. If a point (x, y) is on the circle, it means the coordinates of that point fit perfectly into the equation. Let's tackle the two parts of this problem.
a. Point (-5, a) and the Circle x² + y² = 41
So, we're told the point (-5, a) is on the circle x² + y² = 41. That means if we plug in -5 for 'x' and 'a' for 'y,' the equation should still hold true. Let's do it!
(-5)² + a² = 41 25 + a² = 41 a² = 41 - 25 a² = 16
To find 'a,' we take the square root of both sides. Remember, a square root can have both a positive and a negative value.
a = ±√16 a = ±4
So, the value of 'a' can be either 4 or -4. This means there are two possible points: (-5, 4) and (-5, -4). Both of these points lie on the circle. Pretty neat, huh? We've successfully determined the value of 'a' in this scenario!
b. Point (a, 7) and the Circle x² + y² = 58
Now, let's look at the second part. This time, the point (a, 7) is on the circle x² + y² = 58. We'll follow the same approach, substituting 'a' for 'x' and 7 for 'y'.
a² + 7² = 58 a² + 49 = 58 a² = 58 - 49 a² = 9
Taking the square root of both sides:
a = ±√9 a = ±3
So, in this case, 'a' can be either 3 or -3. This gives us two more points on the circle: (3, 7) and (-3, 7). Awesome! We've successfully figured out the value of 'a' again. The key takeaway is to always substitute the x and y values of the point into the circle's equation and then solve for the unknown variable.
Checking the Locus of Point P(x, y) Relative to Points A and B
Alright, let's switch gears and talk about 'locus.' A locus is a set of all points that satisfy a certain condition. Think of it as the path a point traces as it moves according to a specific rule. We're going to use the distance formula for this because we're dealing with distances between points.
The distance formula is: √[(x₂ - x₁)² + (y₂ - y₁)²]
a. A(-1) and B(0, -25); P(x, y) where PB = 5PA
This one is a bit more involved. We're given points A and B, and we're looking for the path (locus) of point P(x, y) where the distance PB is equal to 5 times the distance PA. This means that the distance from P to B is always five times greater than the distance from P to A. Let's use the distance formula and set up the equation.
First, let's find PA: PA = √[(x - (-1))² + (y - 0)²] PA = √[(x + 1)² + y²]
Next, let's find PB: PB = √[(x - 0)² + (y - (-25))²] PB = √[x² + (y + 25)²]
Now, we know PB = 5PA, so we substitute our distance formulas: √[x² + (y + 25)²] = 5√[(x + 1)² + y²]
To get rid of the square roots, we square both sides: x² + (y + 25)² = 25[(x + 1)² + y²] x² + y² + 50y + 625 = 25(x² + 2x + 1 + y²) x² + y² + 50y + 625 = 25x² + 50x + 25 + 25y²
Now, let's rearrange and simplify this equation. This is where things can get a bit messy, but hang in there. We want to get the equation in a form that we can recognize. Bring everything to one side: 0 = 24x² + 50x + 24y² - 50y - 600
Divide the entire equation by 2 to simplify: 0 = 12x² + 25x + 12y² - 25y - 300
To find the locus, we will need to complete the square for both x and y. Completing the square is a technique used to rewrite quadratic expressions in a more convenient form. This means to transform the equation above into a standard form of a circle.
12(x² + (25/12)x) + 12(y² - (25/12)y) = 300
12(x² + (25/12)x + (25/24)²) + 12(y² - (25/12)y + (25/24)²) = 300 + 12*(25/24)² + 12*(25/24)², where 25/24 = half of the linear term coefficient. We are adding this constant in both sides because of completing the square, adding both sides of the equation by the same constant does not affect the correctness of the equation.
Simplifying and converting it into the standard form of a circle, (x - h)² + (y - k)² = r², where the center is (h, k) and the radius is r.
So, by carefully manipulating the equation, you would eventually find that the locus of point P is a circle. The process will involve completing the square, which I have briefly shown above. By completing the square and simplifying, you would arrive at the equation of a circle.
Therefore, the locus of point P is a circle. Because we used the distance formula, and we know that PB is always 5 times PA, we end up with the equation of a circle.
b. A(1, 1)
This section is incomplete, as the problem statement doesn't give us another point or a condition for the locus. To find the locus, we need a relationship between point P and point A (1, 1). For example, if we were given that PA = k (where k is a constant), the locus would be a circle centered at A with a radius of k. Since this section is incomplete, we cannot fully determine the locus without additional information. We will need more information about the relationship between P(x, y) and A(1, 1) to determine the locus. We need to define some condition, such as a distance relationship, or an angle, to properly define the locus.
Summary
Alright, guys, that was a pretty good run-through of some circle problems! We learned how to find the value of 'a' when a point sits on a circle by plugging its coordinates into the equation and solving for the unknown. We also got a taste of what a locus is – it's the path a point follows based on a specific rule. We saw how to use the distance formula and the importance of using the distance formula. Remember, the locus of a point is determined by its relationship to other points or geometric objects, and this relationship defines its path. When faced with these types of problems, just take them step by step. Identify the given information, set up the equations correctly, and then carefully solve for the unknowns. You've got this!
This guide has hopefully clarified how to approach these kinds of math problems. Keep practicing, and you'll be a pro in no time! Remember to always double-check your calculations, and don't be afraid to ask for help if you get stuck. Good luck, and keep exploring the amazing world of mathematics! Keep in mind the importance of the standard form of a circle equation and how to manipulate equations to reach that format. The core of this topic really is about applying formulas and understanding geometric relationships.