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Question: If the common chord of the circles \(x^{2} + (y - \lambda)^{2} = 16\) and \(x^{2} + y^{2} = 16\) sub...

If the common chord of the circles x2+(y−λ)2=16x^{2} + (y - \lambda)^{2} = 16 and x2+y2=16x^{2} + y^{2} = 16 subtend a right angle at the origin, then λ is equal to

A

4

B

424\sqrt{2}

C

±42\pm 4\sqrt{2}

D

8

Answer

±42\pm 4\sqrt{2}

Explanation

Solution

The common chord of given circles is S1−S2=0(x−2)2+(y−3)2=0S_{1} - S_{2} = 0(x - 2)^{2} + (y - 3)^{2} = 0 i.e., y=λ2y = \frac{\lambda}{2}

(∵λ≠\because\lambda \neq0)

The pair of straight lines joining the origin to the points of intersection of y=λ2y = \frac{\lambda}{2}and x2+y2=16x^{2} + y^{2} = 16 is

x2+y2=16(2yλ)2x^{2} + y^{2} = 16\left( \frac{2y}{\lambda} \right)^{2}

⇒\Rightarrow λ2x2+(λ2−64)y2=0\lambda^{2}x^{2} + (\lambda^{2} - 64)y^{2} = 0.

These lines are at right angles if λ2+λ2−64=0\lambda^{2} + \lambda^{2} - 64 = 0, i.e.,

λ=±42.\lambda = \pm 4\sqrt{2}.