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 x^2 + y^2 - 12x - 4y = 50 that is closest to the origin, we need to minimize the distance between a point on the circle and the origin. This can be done by finding the center of the circle and then determining the shortest distance between the origin and the center.
To find the center of the circle, we rearrange the equation in the standard form:
x^2 - 12x + y^2 - 4y = 50
Completing the square for both x and y terms, we get:
(x^2 - 12x + 36) + (y^2 - 4y + 4) = 50 + 36 + 4
(x - 6)^2 + (y - 2)^2 = 90
By comparing this equation to the standard form of a circle, (x - h)^2 + (y - k)^2 = r^2, we can see that the center is located at (h, k) = (6, 2) and the radius squared is r^2 = 90.
The distance between the origin (0, 0) and the center (6, 2) can be found using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt((6 - 0)^2 + (2 - 0)^2)
d = sqrt(36 + 4)
d = sqrt(40)
d ≈ 6.32
Therefore, the point on the circle that is closest to the origin is the center itself, which is (6, 2).
To find the point on the circle that is farthest from the origin, we need to find the point on the circle that is located at the greatest distance from the center. Since we know the center is (6, 2) and the radius squared is 90, we need to find the point on the circle that lies on the perimeter and is farthest from (6, 2).
We can find this point by finding the tangent to the circle that passes through the center. The line equation of this tangent can be obtained by using the point-slope form:
(y - k) = m(x - h)
Substituting (h, k) = (6, 2) and rearranging, we get:
(y - 2) = m(x - 6)
To find the value of m, we use the fact that the tangent to a circle is perpendicular to the radius at the point of tangency. The slope of the radius is given by:
m_radius = -1/m
Substituting the slope of the tangent into this equation, we get:
m = -1/m_radius
m = -1/0 (as the radius is a vertical line passing through the center)
m = 0
Therefore, the line equation for the tangent is:
y - 2 = 0(x - 6)
y = 2
This equation represents a horizontal line passing through (6, 2), which means that the farthest point on the circle from the origin lies on this line. By substituting y = 2 into the equation of the circle, we can find the x-coordinate of the farthest point:
x^2 + (2^2) - 12x - 4(2) = 50
x^2 + 4 - 12x - 8 = 50
x^2 - 12x - 54 = 0
Solving this quadratic equation, we get two possible values for x:
x ≈ 0.81 and x ≈ 11.19
Therefore, the points on the circle that are farthest from the origin are approximately (0.81, 2) and (11.19, 2).