A point P(x, y) moves in the Cartesian plane such that its distance from the origin O(0, 0… — JAMB UTME

A point $P(x, y)$ moves in the Cartesian plane such that its distance from the origin $O(0, 0)$ is always half of its distance from the fixed point $Q(6, 0)$ . Which of the following equations represents the locus of $P$ ?

$x^2 + y^2 + 4x - 12 = 0$ Answer
B
$x^2 + y^2 - 4x - 12 = 0$
C
$x^2 + y^2 + 12x - 36 = 0$
D
$x^2 + y^2 - 16x + 48 = 0$

Answer

The equation of the locus of

P

x^2 + y^2 + 4x - 12 = 0

The correct equation

x^2 + y^2 + 4x - 12 = 0

is derived by expressing the condition

PO = \frac{1}{2} PQ

2 \cdot PO = PQ

, squaring both sides to get

4(x^2 + y^2) = (x - 6)^2 + y^2

, and simplifying to standard circle form.

Step-by-Step Solution

Express the distance condition algebraically.

The distance from

P(x,y)

O(0,0)

PO = \sqrt{x^2 + y^2}

, and the distance from

P(x,y)

Q(6,0)

PQ = \sqrt{(x-6)^2 + y^2}

. Given

PO = \frac{1}{2} PQ

, we have

2 \cdot PO = PQ

Translate the geometric distance description into algebraic expressions.

Square both sides of the equation to eliminate square roots.

4(x^2 + y^2) = (x - 6)^2 + y^2

Squaring removes the radicals; note that

(2 \cdot PO)^2 = 4 \cdot PO^2

Expand and group like terms.

4x^2 + 4y^2 = x^2 - 12x + 36 + y^2 \implies 3x^2 + 12x + 3y^2 - 36 = 0

Expand

(x - 6)^2 = x^2 - 12x + 36

and collect terms on one side.

Divide the entire equation by 3 to standard form.

x^2 + y^2 + 4x - 12 = 0

Simplifying by the common factor of 3 gives the equation of a circle representing the Circle of Apollonius.

Key Concept

Locus of a point with a constant ratio of distances from two fixed points (Circle of Apollonius)

Estimated Time:2m 0s

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