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
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
Show 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.
Updated On: Jul 9, 2026
- \[ x^2+y^2-3x-4y+\frac94=0 \]
- \[ x^2+y^2+3x+4y+16=0 \]
- \[ x^2+y^2-3x+4y+9=0 \]
- \[ x^2+y^2+3x-4y+\frac94=0 \] \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
If \((h,k)\) is the midpoint of a segment joining a fixed point and a variable point, then the coordinates of the variable point can be expressed in terms of \(h\) and \(k\). Substituting into the given curve gives the locus of the midpoint.
Step 1: Let the midpoint of \(PQ\) be \(M(x,y)\). Let \[ Q=(X,Y). \] Since \(M(x,y)\) is the midpoint of \(PQ\), \[ x=\frac{X+3}{2}, \qquad y=\frac{Y+4}{2}. \] Therefore, \[ X=2x-3, \qquad Y=2y-4. \]
Step 2: Use the fact that \(Q\) lies on the circle. Given \[ X^2+Y^2=16. \] Substituting \[ X=2x-3, \qquad Y=2y-4, \] we get \[ (2x-3)^2+(2y-4)^2=16. \]
Step 3: Expand and simplify. \[ 4x^2-12x+9+4y^2-16y+16=16. \] \[ 4x^2+4y^2-12x-16y+9=0. \] Dividing throughout by \(4\), \[ x^2+y^2-3x-4y+\frac94=0. \]
Step 4: Write the final answer. \[ \boxed{ x^2+y^2-3x-4y+\frac94=0 } \]
Step 1: Let the midpoint of \(PQ\) be \(M(x,y)\). Let \[ Q=(X,Y). \] Since \(M(x,y)\) is the midpoint of \(PQ\), \[ x=\frac{X+3}{2}, \qquad y=\frac{Y+4}{2}. \] Therefore, \[ X=2x-3, \qquad Y=2y-4. \]
Step 2: Use the fact that \(Q\) lies on the circle. Given \[ X^2+Y^2=16. \] Substituting \[ X=2x-3, \qquad Y=2y-4, \] we get \[ (2x-3)^2+(2y-4)^2=16. \]
Step 3: Expand and simplify. \[ 4x^2-12x+9+4y^2-16y+16=16. \] \[ 4x^2+4y^2-12x-16y+9=0. \] Dividing throughout by \(4\), \[ x^2+y^2-3x-4y+\frac94=0. \]
Step 4: Write the final answer. \[ \boxed{ x^2+y^2-3x-4y+\frac94=0 } \]
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