Question

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

• A. $4$
• B. $4\sqrt{2}$
• C. $\pm 4\sqrt{2}$
• D. $8$

Answer 1

The correct option is C $\pm 4\sqrt{2}$

Given equations of circles are $x^2+(y-\lambda {)}^2=16$ and $x^2+y^2=16$
equation of common chord is $S-S^{\prime }=0$
$\Rightarrow 2y-\lambda =0$ -----(1)
$\Rightarrow \frac{2y}{\lambda }=1$
As common chord subtends right angle at the origin,homogenising $x^2+y^2-16=0$
$x^2+y^2-16{\left(\frac{2y}{\lambda }\right)}^2=0$
If $a{x}^2+2hxy+b{y}^2=0$ makes a right angle at origin then $a+b=0$
$\Rightarrow 2-\frac{64}{{\lambda }^2}=0$
$\Rightarrow {\lambda }^2=32$
$\therefore \lambda =\pm 4\sqrt{2}$
Hence, option C.