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 $\pm 4\sqrt{2}$

$C_1:x^2+(y-\lambda {)}^2=16$ & $C_2:x^2+y^2=16$
Chord equation :

$C_1-C_2=0$
$+{\lambda }^2-2\lambda y=0$

$y=\frac{\lambda }{2}$
$x=\pm \sqrt{16-\frac{{\lambda }^2}{4}}$

So length of chord
$=2\sqrt{16-\frac{{\lambda }^2}{4}}$

$\Rightarrow 4^2+4^2=(2\sqrt{16-\frac{{\lambda }^2}{4}}{)}^2$
$16+16=4(16-\frac{{\lambda }^2}{4})$

${\lambda }^2=2\times 16$ $\Rightarrow \lambda =\pm 4\sqrt{2}$