Question
Let $l_{1},l_{2}$ and $l_{3}$ are the lengths of the tangents drawn from a variable point $P$ to the circle $x^{2} + y^{2} = a^{2};x^{2} + y^{2} = 2ax$ and $x^{2} + y^{2} = 2ay$ respectively. The lengths satisfy the relation $l_{1}^{4} = l_{2}^{2}l_{3}^{2} + a^{4}$. Then the locus of $P$ can be:
Options
Solution
[Key Concept: $l_{1},l_{2},l_{3}$ are tangent lengths $\sqrt{S_{i}}$ from $P$ to the three circles; a linear relation among $l_{i}^{2}$ gives the locus of $P$.]
Let P(x, y) be the variable point.
The three circles are:
x² + y² = a²
x² + y² = 2ax
x² + y² = 2ay
Let the lengths of the tangents drawn from P to these circles be l₁, l₂ and l₃ respectively.
For any circle, the square of the tangent length from P is equal to the power of P with respect to the circle.
Therefore,
l₁² = x² + y² − a²
l₂² = x² + y² − 2ax
l₃² = x² + y² − 2ay
Given: l₁⁴ − l₂²l₃² = a⁴
Substituting the above values,
(x² + y² − a²)² − (x² + y² − 2ax)(x² + y² − 2ay) = a⁴
On simplifying,
2a(x + y)(x² + y² − ax − ay) = 0
Therefore, x + y = 0 or x² + y² = a(x + y)
Case 1:
x + y = 0 ,This represents a straight line.
Therefore, (a) Line is possible.
Case 2:
x² + y² = ax + ay, Completing the squares,
x² − ax + y² − ay = 0
(x − a/2)² + (y − a/2)² = a²/2
This represents a circle.
Therefore, (b) Circle is also possible.
Hence, the locus is given by: x + y = 0 or (x − a/2)² + (y − a/2)² = a²/2
(a) Line
(b) Circle
Correct options: (a) and (b)
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