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

Option: 1

4


Option: 2

4 \sqrt{2}


Option: 3

\pm 4 \sqrt{2}


Option: 4

8


Answers (1)

best_answer

The common chord of given circles is \mathrm{S_1-S_2=0}

\mathrm{\Rightarrow x^2+(y-\lambda)^2-16-\left\{x^2+y^2-16\right\}=0 \text { i.e., } y=\frac{\lambda}{2} \quad(\quad \lambda \neq 0)}

The pair of straight lines joining the origin to the points of intersection of  \mathrm{y=\frac{\lambda}{2}} and \mathrm{x^2+y^2=16 \text { is } x^2+y^2=16\left(\frac{2 y}{\lambda}\right)^2}

\mathrm{\Rightarrow \quad \lambda^2 x^2+\left(\lambda^2-64\right) y^2=0} . These lines are at right angles if \mathrm{\lambda^2+\lambda^2-64=0 \text {, i.e., }}

\mathrm{\lambda= \pm 4 \sqrt{2}}

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Ritika Harsh

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