Question

Two circles whose radii are equal to 4 and 8 intersect at right angles. Find the length of their common chord?
(a) $\dfrac{16}{\sqrt{5}}$
(b) 8
(c) $4\sqrt{6}$
(d) $\dfrac{8\sqrt{5}}{5}$

Answer 1

Hint: We start solving the problem by drawing the figure representing all the given information. We then find the distance between the centres of both the circles using the Pythagoras theorem. We then find the sine of angle at one of the centres using its definition. We then use this value for the triangle we get with the common chord as one of its sides.

Complete step-by-step answer:
According to the problem, we are given that the two circles whose radii are equal to 4 and 8 intersect at right angles. We need to find the length of the common chord of those circles.
Let us draw the figure representing the given information.

We know that if two circles intersect orthogonally, then the normal at the points of intersection are perpendicular. i.e., the angle at the point C is ${{90}^{\circ }}$ (as we know that the normal passes through the centre of the circle).
From the Pythagoras theorem, we know that the square of the hypotenuse is equal to the sum of the squares of the other two sides. We apply this to triangle ABC.
So, we get ${{d}^{2}}={{8}^{2}}+{{4}^{2}}$.
$\Rightarrow {{d}^{2}}=64+16$.
$\Rightarrow {{d}^{2}}=80$.
$\Rightarrow d=\sqrt{80}$.
$\Rightarrow d=4\sqrt{5}$ ---(1).
Let us find the angle at point B. We know that sine of an angle in a right-angled triangle is defined as the ratio of opposite sides to the hypotenuse.
So, we get $\sin B=\dfrac{4}{4\sqrt{5}}$.
$\Rightarrow \sin B=\dfrac{1}{\sqrt{5}}$ ---(2).
From the figure, we can see that the triangle CFB is a right-angles triangle with right angle at vertex F.
We get $\sin B=\dfrac{x}{8}$.
From equation (2), we get $\dfrac{1}{\sqrt{5}}=\dfrac{x}{8}$.
$\Rightarrow x=\dfrac{8}{\sqrt{5}}$ ---(3).
But we need to find the length of the common chord which is CD. From the figure, we can see that the length of CD is twice the length of the CF.
So, we get the length of the common chord = $2\times \dfrac{8}{\sqrt{5}}=\dfrac{16}{\sqrt{5}}$.
∴ The length of the common chord is $\dfrac{16}{\sqrt{5}}$.

So, the correct answer is “Option A”.

Note: We should know that the common chord intersects the line segment joining the centres perpendicularly. We should not make calculation mistakes while solving this problem. We should not stop solving the problem after finding the value of x as we need the length of the common chord. We can also use the area of the triangle by assuming different sides as base of the triangle (as CF is the altitude in the triangle ABC). Similarly, we can expect problems to find the length of the common chords.