Which point on the circle x^2+y^2-12x-4y=50 is closest to the origin? Which point is farthest from the origin? Explain.
To find the point on the circle that is closest to the origin, we need to minimize the distance between the origin and any point on the circle. Similarly, to find the point on the circle that is farthest from the origin, we need to maximize this distance.
The equation of the circle x^2 + y^2 - 12x - 4y = 50 can be rewritten in the standard form as (x - 6)^2 + (y - 2)^2 = 114. Let's analyze the expression inside the standard form: (x - 6)^2 + (y - 2)^2.
From the equation, we can see that the center of the circle is at the point (6, 2) because the equation represents the distance between each point on the circle and the center. The distance of any point on the circle from the origin can be found using the distance formula.
The distance between the origin (0, 0) and any point (x, y) on the circle can be calculated as:
d = √[(x - 0)^2 + (y - 0)^2] = √[x^2 + y^2]
To find the point on the circle closest to the origin, we need to minimize the distance expression. Since x^2 and y^2 are both non-negative, the sum of x^2 + y^2 is minimized when both x^2 = 0 and y^2 = 0. Therefore, the point on the circle closest to the origin is the point (6, 2), which is the center of the circle.
To find the point on the circle farthest from the origin, we need to maximize the distance expression. Since x^2 and y^2 are non-negative, the maximum value of x^2 + y^2 is obtained when one of them is at its maximum. In this case, the maximum value of x^2 + y^2 can be found at the farthest point on the circle from the center, which coincides with the farthest point on the circle from the origin. Since the distance is determined by the radius of the circle, the farthest point will lie on the circumference of the circle.
The radius of the circle is √114. To find the farthest point, we need to find a point on the circumference of the circle that is √114 units away from the center (6, 2). Let's call this point (x', y').
Using the distance formula, we have:
√[(x' - 6)^2 + (y' - 2)^2] = √114
Squaring both sides of the equation, we have:
(x' - 6)^2 + (y' - 2)^2 = 114
Since we aim to find the farthest point, we can maximize this equation. The farthest point from the origin will be the solution for (x', y').
Simplifying the equation further, we get:
(x' - 6)^2 + (y' - 2)^2 = 114
This equation represents an equation of a circle centered at (6, 2) with a radius of √114.
Therefore, the farthest point from the origin lies on the circumference of this circle.