Question

If the chord of contact of tangents drawn from a point $(\alpha ,\beta )$ to the circle $x^2+y^2=a^2$ subtends a right angle at the centre of the circle, then:

• A. ${\alpha }^2+{\beta }^2=a^2$
• B. ${\alpha }^2+{\beta }^2=2a^2$
• C. ${\alpha }^2+{\beta }^2=2/a^2$
• D. ${\alpha }^2+{\beta }^2=a^2/2$

Answer 1

Answer: (B)

${\alpha }^2+{\beta }^2=2a^2$

The given equation of circle is,

$x^2+y^2=a^2$

$\therefore {(x-0)}^2+{(y-0)}^2=a^2$

Comparing this equation with standard form i.e. ${(x-h)}^2+{(y-k)}^2=r^2$, we get

$h=0$, $k=0$ and $r=a$

Thus, Center of circle is at origin i.e. $O(0,0)$ and $OA=OB=a$

Now, AC is tangent to the circle through point A.

$\therefore \angle CAO={90}^0$

Similarly, BC is tangent to circle to a circle through point B.

$\therefore \angle CBO={90}^0$

Also, $\angle AOB={90}^0$ (Given)

Thus, we can conclude that $\angle ACB={90}^0$

Similarly, $OA=OB=AC=BC$

Thus, $\square AOBC$ is square.

Now, by distance formula, $OC=\sqrt{{(\alpha -0)}^2+{(\beta -0)}^2}$

$\therefore OC=\sqrt{{\alpha }^2+{\beta }^2}$

By property of square, diagonals are equal in length.

$\therefore AB=OC=\sqrt{{\alpha }^2+{\beta }^2}$

Now, in right angled triangle $\Delta AOB$, by Pythagoras theorem,

${AB}^2={OA}^2+{OB}^2$

$\therefore {\left(\sqrt{{\alpha }^2+{\beta }^2}\right)}^2=a^2+a^2$

$\therefore {\alpha }^2+{\beta }^2={2a}^2$

Thus, answer is option (B)