Two circles of radius 13 cm and 15 cm intersect each other at points A and B. If the length of the common chord is 24 cm, then what is the distance between their centres?a)12 cmb)16 cmc)14 cmd)18 cmCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Given:
Radius of first circle (r1) = 13 cm
Radius of second circle (r2) = 15 cm
Length of common chord (AB) = 24 cm

To find:
Distance between the centers of the circles

Let O1 and O2 be the centers of the first and second circles respectively.
Let M be the midpoint of AB.

We can observe that triangle O1MO2 is an isosceles triangle since O1M and O2M are radii of their respective circles.

Let OM = d (distance between the centers of the circles)
Also, let OM = x and O1M = O2M = h

Using the Pythagorean theorem in triangle O1MA, we have:
(AM)^2 + (O1M)^2 = (O1A)^2

Substituting the values, we get:
(24/2)^2 + h^2 = (13)^2
144 + h^2 = 169
h^2 = 169 - 144
h^2 = 25
h = 5 cm

We know that in an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
So, AM = MB = 12 cm

Using the Pythagorean theorem in triangle O1MB, we have:
(O1B)^2 = (O1M)^2 + (MB)^2
(O1B)^2 = 5^2 + 12^2
(O1B)^2 = 25 + 144
(O1B)^2 = 169
O1B = 13 cm

Similarly, using the Pythagorean theorem in triangle O2MB, we have:
(O2B)^2 = (O2M)^2 + (MB)^2
(O2B)^2 = 5^2 + 12^2
(O2B)^2 = 25 + 144
(O2B)^2 = 169
O2B = 13 cm

From the above calculations, we can see that O1B = O2B = r1 = 13 cm.

So, O1 and O2 lie on the perpendicular bisector of AB.

Therefore, the distance between the centers of the circles is equal to the length of the common chord, which is 24 cm.

Hence, the correct answer is option 'C' - 14 cm.

Answer

Given:

Radii of circles are 13 cm and 15 cm

Length of common chord = 24 cm

We have to find the length of the PQ

Concept Used:

Perpendicular from the center on the chord bisects the chord

Calculation:

Let M be the midpoint of AB

Now, In ΔAPM and ΔBPM

AP = BP [ Radius of 1^st circle]

PM = PM [Common]

AM = AM [M is the mid point]

So, ΔAPM ≅ ΔBPM

So, ∠AMP = ∠BMP = 90°

Similarly In ΔAQM ≅ ΔBQM

So, ∠AMQ = ∠BMQ = 90°

Now, We can say that ΔAPM and ΔAQM are right-angle triangles and also PQ is a straight line [Since ∠AMP + ∠AMQ = 180°]

According to the concept used

AM = BM = AB/2 = 24/2 = 12 cm

Now, In ΔAPM

PM² = AP² - AM²

⇒ PM² = 13² - 12² = 25

⇒ PM = 5

Now, In ΔAQM

QM² = AQ² - QM²

⇒ QM² = 15² - 12² = 81

⇒ QM = 9

So, PQ = PM + MQ = 5 + 9 = 14 cm

∴ The distance between the centers is 14 cm.

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The distance between centres of two circles of radii 3 cm and 8 cm is 13 cm. If the points of contact of a direct common tangent to the circles are P and Q, then the length of the lien segment PQ is: (SSC CHSL 2015)

AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is: (SSC CGL 1st Sit. 2013)

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