Conic Sections-

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.

To show that the locus is a circle, we need to find its equation, center, and radius.

Let's assume the coordinates of the moving point are (x, y). Then, according to the given condition, the distance between the point (x, y) and the origin (0, 0) is twice the distance between the point (x, y) and the point (3, 0).

Using the distance formula, the first condition can be written as:

√(x² + y²) = 2√((x - 3)² + y²)

Let's square both sides of the equation to eliminate the square root:

x² + y² = 4(x - 3)² + 4y²

Expanding the equation, we get:

x² + y² = 4(x² - 6x + 9) + 4y²

x² + y² = 4x² - 24x + 36 + 4y²

Rearranging the equation, we get:

3x² + 24x - 36 = 0

Now, we have a quadratic equation in standard form. To determine its shape and find its center and radius, we can write it in general form (Ax² + Bxy + Cy² + Dx + Ey + F = 0) and analyze its coefficients:

3x² + 24x - 36 = 0
Divide the entire equation by 3 to simplify:
x² + 8x - 12 = 0

Comparing this with the general form (Ax² + Bxy + Cy² + Dx + Ey + F = 0), we have:
A = 1, B = 0, C = 0, D = 8, E = 0, F = -12

Since B and C are both zero, we can conclude that the equation does not have a cross-term (xy term), indicating that the conic section is a circle.

To find the center and radius of the circle, we can complete the square:

x² + 8x + 16 - 16 - 12 = 0
(x + 4)² - 28 = 0
(x + 4)² = 28

Now, we have the equation of a circle in the form (x - h)² + (y - k)² = r², where (h, k) represents the center of the circle and r represents the radius.

Comparing our equation, we have:
(x + 4)² = 28

So, the center of the circle is (-4, 0), and the radius can be found by taking the square root of the constant term:

r = √28 = 2√7

Therefore, the locus of the moving point is a circle with center (-4, 0) and radius 2√7.