Question

If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is $90^{\circ}$, then the length (in cm) of their common chord is :

• A. $\frac{60}{13}$
• B. $\frac{120}{13}$
• C. $\frac{13}{2}$
• D. $\frac{13}{5}$

Answer 1

The Correct Option is B

To find the length of the common chord of two circles intersecting at a \(90^{\circ}\) angle, we can use the following geometrical property:

Let the two circles with centers \(O_1\) and \(O_2\) have radii \(r_1\) and \(r_2\) respectively. When the angle between the tangents at their points of intersection is \(90^{\circ}\), the distance between the centers \(O_1\) and \(O_2\) is \(\sqrt{r_1^2 + r_2^2}\).

In this problem:

• The radius of the first circle \(r_1 = 5 \text{ cm}\).
• The radius of the second circle \(r_2 = 12 \text{ cm}\).

The distance \(d\) between the centers of the two circles is:

\(d = \sqrt{r_1^2 + r_2^2}\)

\(d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}\)

Let \(AB\) be the common chord, and let \(M\) be the midpoint of \(AB\). The relationship between the circles, the chord, and the distance is given by the formula:

\(l = 2 \sqrt{r_1^2 - \left(\frac{d^2 - r_2^2 + r_1^2}{2d}\right)^2}\)

Where \(l\) is the length of the common chord. Substituting the given values:

The perpendicular distance from the midpoint of the chord to the center of the circle that can be represented as \(h\) is given by:

\(h = \frac{d^2 - r_2^2 + r_1^2}{2d} = \frac{13^2 - 12^2 + 5^2}{2 \times 13}\)

\(h = \frac{169 - 144 + 25}{26} = \frac{50}{26} = \frac{25}{13} \text{ cm}\)

Now substituting \(h\) into the formula for the chord:

\(l = 2 \sqrt{r_1^2 - h^2} = 2 \sqrt{5^2 - \left(\frac{25}{13}\right)^2}\)

\(l = 2 \sqrt{25 - \frac{625}{169}}\)

\(l = 2 \sqrt{\frac{4225}{169} - \frac{625}{169}}\)

\(l = 2 \sqrt{\frac{3600}{169}} = 2 \times \frac{60}{13} = \frac{120}{13} \text{ cm}\)

Therefore, the length of the common chord is \(\frac{120}{13}\) cm.