Let L1, L2 and L3 be the lengths of tangents drawn from a point P (h, k) to the circles x2 + y2 = 4, x2 + y2 – 4x = 0 and x2 + y2 – 4y = 0 respectively. If L14 = L22 L32 + 16 then locus of P are the curves, C1 (straight line) and C2 (circle).

Let L₁, L₂ and L₃ be the lengths of tangents drawn from a point P (h, k) to the circles x² + y² = 4, x² + y² – 4x = 0 and x² + y² – 4y = 0 respectively. If L₁⁴ = L₂² L₃² + 16 then locus of P are the curves, C₁ (straight line) and C₂ (circle).

Linked Question 1

Straight line which intersects both the curves C₁ and C₂ orthogonally, is

x – y = 0

x + y = 0

x – y – 2 = 0

x + y + 2 = 0

$Clearly, L_{1} = \sqrt{h^{2} + k^{2} - 4}, L_{2} = \sqrt{h^{2} + k^{2} - 4 h}, L_{3} = \sqrt{h^{2} + k^{2} - 4 k}$

Now, L₁⁴ = L₂² L₃² + 16 (Given)
⟹ (h² + k² – 4)² = (h² + k² – 4h) (h² + k² – 4k) + 16
⟹ (h² + k²)² – 8(h² + k²) + 16 = (h² + k²)² – 4(h² + k²) (h + k) + 16hk + 16
⟹ 8(h + k)² – 4(h² + k²) (h + k) = 0
⟹ 4 (h + k) [h² + k² – 2(h + k)] = 0
So, locus of P (h, k) are
C₁ : x + y = 0 and C₂ : x² + y² – 2x – 2y = 0

From the figure it is clear that straight line
which intersects both the curves orthogonally is, x – y = 0.

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Linked Question 2

If S₁, S₂ and S₃ are three circles congruent to C₂ and touches both the curves C₁ and C₂, then the area of triangle formed by joining centres of the circles S₁, S₂ and S₃, is

$Clearly, L_{1} = \sqrt{h^{2} + k^{2} - 4}, L_{2} = \sqrt{h^{2} + k^{2} - 4 h}, L_{3} = \sqrt{h^{2} + k^{2} - 4 k}$

$Area of the ∆ ABC = \frac{1}{2} \times (4 r) \times (2 r) = 4 r^{2} = 4 (\sqrt{2})^{2} = 8$

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Linked Question 3

Circumcentre of the triangle formed by C₁ and two other lines which are at an angle of 45° with C₁ and tangent to the circle C₂, is

(–1, –1)

(1, 1)

(2, 2)

(0, 0)

$Clearly, L_{1} = \sqrt{h^{2} + k^{2} - 4}, L_{2} = \sqrt{h^{2} + k^{2} - 4 h}, L_{3} = \sqrt{h^{2} + k^{2} - 4 k}$

Δ PQR is required triangle which is right angled isosceles
triangle and its circumcentre is (0, 0).

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