A point P(x, y) moves in the Cartesian plane such that it maintains a constant distance of… — JAMB UTME

A point $P(x, y)$ moves in the Cartesian plane such that it maintains a constant distance of $10\text{ units}$ from a fixed point $C(2, -3)$ . If the locus of $P$ intersects the vertical line $x = 8$ at two points $A$ and $B$ , what is the distance between $A$ and $B$ ?

Answer: 16 units

Answer

The distance between the intersection points A and B is 16 units.

The locus of point P moving at a constant distance of 10 units from C(2, -3) forms a circle

(x - 2)^2 + (y + 3)^2 = 100

. Substituting

x = 8

yields

(y + 3)^2 = 64

, giving

y = 5

and

y = -11

. The distance between the two points (8, 5) and (8, -11) along the vertical line

x = 8

5 - (-11) = 16

units.

Step-by-Step Solution

Determine the equation representing the locus of point P

(x - 2)^2 + (y + 3)^2 = 100

The locus of a point moving at a fixed distance from a fixed point is a circle with equation

(x - h)^2 + (y - k)^2 = r^2

Substitute

x = 8

into the locus equation to find the

y

-coordinates of the intersection points

(8 - 2)^2 + (y + 3)^2 = 100 \implies 36 + (y + 3)^2 = 100 \implies (y + 3)^2 = 64

The intersection points lie on both the locus circle and the vertical line

x = 8

Solve for the two possible values of

y

y + 3 = \pm 8 \implies y_1 = 5

and

y_2 = -11

Taking the square root gives both positive and negative solutions for the vertical coordinate.

Calculate the vertical distance between points

A(8, 5)

and

B(8, -11)

\text{Distance} = 5 - (-11) = 16\text{ units}

Since both points have the same

x

-coordinate (

x = 8

), the distance is simply the absolute difference between their

y

-coordinates.

Key Concept

Locus of a point at a constant distance from a fixed point (Circle)

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