a point moves so that its distance from the origin is twice its distance from the point (3,0) show that the locus is a circle and find its centre and radius
let the point be (x,y)
(√(x^2 + y^2) = 2√(x-3)^2 + y^2) = 0
square both sides
x^2 + y^2 = 4(x^2 - 6x + 9) + y^2)
x^2 + y^2 = 4x^2 - 24x + 36 + 4y^2
3x^2 + 3y^2 - 24x + 36 = 0
x^2 + y^2 - 8x + 12 = 0
complete the square
x^2 - 8x + 16 + y^2 = -12 + 16
(x - 4)^2 + y^2 = 4
indeed a circle with centre (4,0) and radius 2
To begin, let's assume the coordinates of the moving point are (x, y). The distance of this point from the origin (0, 0) can be determined using the distance formula:
Distance from origin = √(x^2 + y^2)
The distance of the point from the point (3, 0) can be found using the distance formula as well:
Distance from (3, 0) = √((x-3)^2 + y^2)
According to the given condition, the distance from the origin is twice the distance from the point (3, 0). Therefore, we can write the equation as:
√(x^2 + y^2) = 2 * √((x-3)^2 + y^2)
To simplify, we square both sides of the equation:
(x^2 + y^2) = 4 * ((x-3)^2 + y^2)
Expanding the equation:
x^2 + y^2 = 4(x^2 - 6x + 9 + y^2)
Simplifying further:
x^2 + y^2 = 4x^2 - 24x + 36 + 4y^2
Rearranging the equation to express it in the standard form of a circle (x^2 + y^2 + Dx + Ey + F = 0):
3x^2 - 24x + 3y^2 - 36 = 0
Now, let's complete the square for both x and y terms to put it in the general form of a circle equation:
3(x^2 - 8x) + 3(y^2 - 12) = 36
3(x^2 - 8x + 16) + 3(y^2 - 12) = 36 + 3(16)
3(x - 4)^2 + 3(y^2 - 12) = 84
Dividing both sides of the equation by 3:
(x - 4)^2 + (y^2 - 12) = 28
Recalling that the general equation of a circle with the center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = r^2, we can identify the center and radius of the locus.
Comparing the given equation with the general equation, we find that the center of the locus is (h, k) = (4, 0) and the radius squared is r^2 = 28. Therefore, the radius of the locus is r = √28 or 2√7.
Hence, the locus is a circle centered at (4, 0) with a radius of 2√7.
To start, let's assume the coordinates of the point are (x, y). We can find the distance of the point from the origin (0, 0) using the distance formula:
Distance from origin = √((x - 0)^2 + (y - 0)^2) = √(x^2 + y^2)
Similarly, we can find the distance of the point from the point (3, 0) using the distance formula:
Distance from (3, 0) = √((x - 3)^2 + (y - 0)^2) = √((x - 3)^2 + y^2)
According to the problem, the distance from the origin is twice the distance from the point (3, 0). Mathematically, we can express this as:
√(x^2 + y^2) = 2 * √((x - 3)^2 + y^2)
Let's square both sides to eliminate the square root:
(x^2 + y^2) = 4 * ((x - 3)^2 + y^2)
Expanding and simplifying:
x^2 + y^2 = 4 * (x^2 - 6x + 9 + y^2)
x^2 + y^2 = 4x^2 - 24x + 36 + 4y^2
Simplifying further:
3x^2 + 24x - 4y^2 - 36 = 0
Rearranging the terms:
3x^2 + 24x - (4y^2 + 36) = 0
Dividing by 3 to make the coefficient of x^2 equal to 1:
x^2 + 8x - (4y^2 + 36)/3 = 0
Now, we can complete the square to express the equation in the standard form of a circle:
x^2 + 8x + (8/2)^2 = (4y^2 + 36)/3 + (8/2)^2
x^2 + 8x + 16 = (4y^2 + 36)/3 + 16
(x + 4)^2 = (4y^2 + 36)/3 + 48/3
(x + 4)^2 = (4y^2 + 36 + 48)/3
(x + 4)^2 = (4y^2 + 84)/3
Multiplying both sides by 3 to eliminate the fraction:
3 * (x + 4)^2 = 4y^2 + 84
This equation represents a circle since variables x and y are squared, and their coefficients are the same. The equation is now in the standard form of a circle:
(x + 4)^2 + (4y^2 + 84)/4 = 1
Comparing this equation with the general equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
We can determine that the center of the circle is (-4, 0), and the radius of the circle is √(4y^2 + 84)/2 or √(y^2 + 21).