If two chords of lengths $2 a$ each, of a circle of radius $R$ , intersect each other at right angles then the distance of their point of intersection from the centre of the circle is

2 \sqrt{R^{2} - a^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2 (R^{2} - a^{2})}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 \sqrt{(R^{2} - a^{2})}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (R^{2} - a^{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{2 (R^{2} - a^{2})}$
In quadrilateral $O N A M$
$∠ N = 90^{\circ}, ∠ M = 90^{\circ}$
$∠ A = 90^{\circ}$ , so $∠ O = 90^{\circ}$
$O M = O N = \sqrt{R^{2} - a^{2}}$
So, $O N A M$ is a square
Diagonal of square
$O N A M = O A$
$= \sqrt{2} \sqrt{R^{2} - a^{2}}$
$= \sqrt{2 (R^{2} - a^{2})}$

Suggest Corrections

Similar questions

r (< a)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)

(a,b) is the mid point of the chord $¯ A B$ of the circle $x^{2} + y^{2} = r^{2}$ . The tangent at A,B meet a C. then area of $Δ A B C$

a^{2} - 1

3 (a + 1)

\frac{15}{2}

x^{2} + y^{2} = a^{2}

Join BYJU'S Learning Program