Question:
A point \(P(x,y)\) is such that its distances from \((-1,0)\) and \((0,2)\) are in the ratio of
\[
\sqrt{2}:1
\]
Then the locus of \(P\) is
A point \(P(x,y)\) is such that its distances from \((-1,0)\) and \((0,2)\) are in the ratio of
\[
\sqrt{2}:1
\]
Then the locus of \(P\) is
Show Hint
When distances of a moving point from two fixed points are given in a constant ratio, first use the distance formula and then simplify the equation by completing the squares to identify the locus.
Updated On: Jun 25, 2026
- \((x-1)^2+(y-4)^2=10\)
- \((x+2)^2+(y+2)^2=10\)
- \((x-1)^2+(y-4)^2=100\)
- \((x+2)^2+(y+2)^2=100\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the distance condition.
Let \[ A(-1,0) \] and \[ B(0,2) \] Given that the distances of \(P(x,y)\) from \(A\) and \(B\) are in the ratio \[ \sqrt2:1 \] Therefore, \[ \frac{PA}{PB}=\sqrt2 \] Squaring both sides, \[ \frac{PA^2}{PB^2}=2 \]
Step 2: Use the distance formula.
Distance from \(A(-1,0)\): \[ PA^2=(x+1)^2+y^2 \] Distance from \(B(0,2)\): \[ PB^2=x^2+(y-2)^2 \] Using the ratio condition, \[ (x+1)^2+y^2 = 2\left[x^2+(y-2)^2\right] \]
Step 3: Expand and simplify.
Expanding the left side, \[ x^2+2x+1+y^2 \] Expanding the right side, \[ 2(x^2+y^2-4y+4) \] \[ = 2x^2+2y^2-8y+8 \] Therefore, \[ x^2+2x+1+y^2 = 2x^2+2y^2-8y+8 \] Bringing all terms to one side, \[ 0 = x^2+y^2-2x-8y+7 \] So, \[ x^2-2x+y^2-8y=-7 \]
Step 4: Complete the squares.
Completing square in \(x\), \[ x^2-2x=(x-1)^2-1 \] Completing square in \(y\), \[ y^2-8y=(y-4)^2-16 \] Substituting, \[ (x-1)^2-1+(y-4)^2-16=-7 \] Thus, \[ (x-1)^2+(y-4)^2=10 \]
Step 5: Final conclusion.
Hence, the locus of \(P\) is \[ \boxed{(x-1)^2+(y-4)^2=10} \]
Let \[ A(-1,0) \] and \[ B(0,2) \] Given that the distances of \(P(x,y)\) from \(A\) and \(B\) are in the ratio \[ \sqrt2:1 \] Therefore, \[ \frac{PA}{PB}=\sqrt2 \] Squaring both sides, \[ \frac{PA^2}{PB^2}=2 \]
Step 2: Use the distance formula.
Distance from \(A(-1,0)\): \[ PA^2=(x+1)^2+y^2 \] Distance from \(B(0,2)\): \[ PB^2=x^2+(y-2)^2 \] Using the ratio condition, \[ (x+1)^2+y^2 = 2\left[x^2+(y-2)^2\right] \]
Step 3: Expand and simplify.
Expanding the left side, \[ x^2+2x+1+y^2 \] Expanding the right side, \[ 2(x^2+y^2-4y+4) \] \[ = 2x^2+2y^2-8y+8 \] Therefore, \[ x^2+2x+1+y^2 = 2x^2+2y^2-8y+8 \] Bringing all terms to one side, \[ 0 = x^2+y^2-2x-8y+7 \] So, \[ x^2-2x+y^2-8y=-7 \]
Step 4: Complete the squares.
Completing square in \(x\), \[ x^2-2x=(x-1)^2-1 \] Completing square in \(y\), \[ y^2-8y=(y-4)^2-16 \] Substituting, \[ (x-1)^2-1+(y-4)^2-16=-7 \] Thus, \[ (x-1)^2+(y-4)^2=10 \]
Step 5: Final conclusion.
Hence, the locus of \(P\) is \[ \boxed{(x-1)^2+(y-4)^2=10} \]
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