Question

If \(P(3,4)\) is a fixed point and \(Q\) is a variable point on the circle \[ x^2+y^2=16, \] then the locus of the midpoint of \(PQ\) is

• A. \[ x^2+y^2-3x-4y+\frac94=0 \]
• B. \[ x^2+y^2+3x+4y+16=0 \]
• C. \[ x^2+y^2-3x+4y+9=0 \]
• D. \[ x^2+y^2+3x-4y+\frac94=0 \] \bigskip

Hint:

For midpoint-locus problems, let the midpoint be \((x,y)\). Express the coordinates of the variable endpoint using the midpoint formula and substitute into the given curve.

Answer 1

The Correct Option is A

Concept: Use the midpoint formula parametrically. Let the variable point Q on the circle be \((4\cos\theta, 4\sin\theta)\). Midpoint M of P(3,4) and Q is \(\left(\frac{3+4\cos\theta}{2}, \frac{4+4\sin\theta}{2}\right)\). Eliminate \(\theta\) to find the locus.

Step 1: Express M coordinates. \(x = \frac{3+4\cos\theta}{2}, y = \frac{4+4\sin\theta}{2}\).

Step 2: Solve for \(\cos\theta, \sin\theta\). \(\cos\theta = \frac{2x-3}{4}, \sin\theta = \frac{2y-4}{4}\).

Step 3: Use \(\cos^2\theta + \sin^2\theta = 1\). \(\left(\frac{2x-3}{4}\right)^2 + \left(\frac{2y-4}{4}\right)^2 = 1 \Rightarrow (2x-3)^2 + (2y-4)^2 = 16\).

Step 4: Expand. \(4x^2-12x+9+4y^2-16y+16=16 \Rightarrow 4x^2+4y^2-12x-16y+9=0\). Divide by 4: \(x^2+y^2-3x-4y+\frac94=0\).

Step 5: Write the final answer. \(\boxed{x^2+y^2-3x-4y+\frac94=0}\)