Question

The line joining \((5,0)\) to \((10\cos\theta,10\sin\theta)\) is divided internally in the ratio \(2:3\) at \(P\). If \(\theta\) varies, then the locus of \(P\) is

• A. a straight line
• B. a pair of straight lines
• C. a circle
• D. None of the above

Hint:

When one point moves on a circle and another point is fixed, a point dividing the segment in a constant ratio also traces a circle (scaled and shifted).

Answer 1

The Correct Option is C

Step 1: Points involved.
Let:
\[ A(5,0),\quad B(10\cos\theta,10\sin\theta) \] Point \(P\) divides \(AB\) internally in ratio \(2:3\).
Step 2: Use section formula.
If \(AP:PB=2:3\), then:
\[ P\left(\frac{3x_1+2x_2}{5},\frac{3y_1+2y_2}{5}\right) \] Here:
\[ (x_1,y_1)=(5,0),\quad (x_2,y_2)=(10\cos\theta,10\sin\theta) \] So:
\[ x=\frac{3(5)+2(10\cos\theta)}{5} =\frac{15+20\cos\theta}{5} =3+4\cos\theta \] \[ y=\frac{3(0)+2(10\sin\theta)}{5} =\frac{20\sin\theta}{5} =4\sin\theta \] Step 3: Eliminate \(\theta\).
\[ x-3=4\cos\theta,\quad y=4\sin\theta \] Square and add:
\[ (x-3)^2+y^2=16(\cos^2\theta+\sin^2\theta)=16 \] Step 4: Identify locus.
\[ (x-3)^2+y^2=16 \] This is a circle with centre \((3,0)\) and radius 4.
Final Answer: \[ \boxed{\text{a circle}} \]