The Locus of a Point and Circle Geometry: Exploring the Distance from Two Fixed Points

In this article, we will explore the mathematical principles behind finding the locus of a point.

The problem at hand is to find the locus of a point such that its distance from point P(2, 0) is always twice its distance from point Q(-1, 3). Let's dive into the detailed steps and algebraic manipulations to reach our solution.

Understanding the Problem

The problem is to find a point (x, y) whose distance from P(2, 0) is always twice its distance from Q(-1, 3). Let's set up the equation based on the distance formula.

The distance from (x, y) to P(2, 0) is: [ d_{RP} sqrt{(x - 2)^2 y^2} ] The distance from (x, y) to Q(-1, 3) is: [ d_{RQ} sqrt{(x 1)^2 (y - 3)^2} ]

Step 1: According to the problem, the distance from (x, y) to P(2, 0) is twice its distance to Q(-1, 3). Therefore, we can write the equation:

[ sqrt{(x - 2)^2 y^2} 2 cdot sqrt{(x 1)^2 (y - 3)^2} ]

Step 2: Square both sides to eliminate the square roots:

[ (x - 2)^2 y^2 4 cdot ((x 1)^2 (y - 3)^2) ]

Algebraic Manipulation

Let's expand and simplify this equation step by step.

Step 3: Expand the left side:

[ (x - 2)^2 y^2 x^2 - 4x 4 y^2 ]

Step 4: Expand the right side:

[ 4 cdot ((x 1)^2 (y - 3)^2) 4(x^2 2x 1 y^2 - 6y 9) 4x^2 8x 4 4y^2 - 24y 36 4x^2 4y^2 8x - 24y 40 ]

Step 5: Substitute these expansions back into the equation and combine like terms:

[ x^2 - 4x 4 y^2 4x^2 4y^2 8x - 24y 40 ] [ x^2 - 4x y^2 4x^2 4y^2 8x - 24y 40 ] [ 0 3x^2 3y^2 12x - 24y 36 ] [ x^2 y^2 4x - 8y 12 0 ]

Step 6: Complete the square for both x and y:

For x: [ x^2 4x (x 2)^2 - 4 ] For y: [ y^2 - 8y (y - 4)^2 - 16 ]

Step 7: Substitute these into the equation:

[ (x 2)^2 - 4 (y - 4)^2 - 16 12 0 ] [ (x 2)^2 (y - 4)^2 - 8 0 ] [ (x 2)^2 (y - 4)^2 8 ]

This represents a circle with center (-2, 4) and radius sqrt{8} 2sqrt{2}.

General Form and Solution Verification

Let's extend the solution to a more general case.

We are now going to find the locus of points whose distance from P(a, b) is twice the distance from Q(c, d). Consider the general point (x, y). The distance from (x, y) to P(a, b) is:

[ d_{AP} sqrt{(x - a)^2 (y - b)^2} ]

The distance from (x, y) to Q(c, d) is:

[ d_{AQ} sqrt{(x - c)^2 (y - d)^2} ]

From the given condition, we have:

[ sqrt{(x - a)^2 (y - b)^2} 2 cdot sqrt{(x - c)^2 (y - d)^2} ]

Squaring both sides and expanding the brackets, we obtain:

[ (x - a)^2 (y - b)^2 4 cdot ((x - c)^2 (y - d)^2) ]

This simplifies to:

[ x^2 - 2ax a^2 y^2 - 2by b^2 4(x^2 - 2cx c^2 y^2 - 2dy d^2) ]

Further simplification yields:

[ x^2 - 2ax y^2 - 2by a^2 b^2 4x^2 - 8cx 4c^2 4y^2 - 8dy 4d^2 ]

Combining like terms and setting everything to zero:

[ 3x^2 3y^2 - 2ax - 8cx 8cx - 2by 8dy - 4c^2 - 4d^2 - a^2 - b^2 0 ]

Dividing through by 3:

[ x^2 y^2 - frac{2a - 8c}{3}x - frac{2b - 8d}{3}y - frac{4c^2 4d^2 - a^2 - b^2}{3} 0 ]

Completing the square:

[ left(x frac{2a - 8c}{6}right)^2 - left(frac{2a - 8c}{6}right)^2 left(y - frac{2b - 8d}{6}right)^2 - left(frac{2b - 8d}{6}right)^2 frac{4c^2 4d^2 - a^2 - b^2}{3} ]

Thus:

[ left(x frac{2a - 8c}{6}right)^2 left(y - frac{2b - 8d}{6}right)^2 frac{4}{9}(c^2 d^2 a^2 b^2) ]

Therefore, the locus of points is a circle with center:

[ left(-frac{2a - 8c}{6}, frac{2b - 8d}{6}right) ]

And radius:

[ frac{2}{3} sqrt{a^2 b^2 - 4c^2 - 4d^2} ]

Substituting a 2, b 0, c -1, d 3, we have:

[ text{Center} left(-frac{4 - 8}{6}, frac{0 - 24}{6}right) (-2, 4) ] [ text{Radius} frac{2}{3} sqrt{4 9 - 4 - 36} 2sqrt{2} ]

So, the final answer is:

The locus of the point is a circle with center (-2, 4) and radius 2sqrt{2}.