Question

Two circles of radius 5 cm and 3 cm intersect a two points and the distance between their centres is 4cm . find the lenght of the common chord.​

Answer 1

Answer:

hey mate ,

Step-by-step explanation:

→ solution : Let PQ be the common chord of the two circles having their centres at c₁ and c₂.

→ We have ,  

→ 5² = 4 ² + 3²

→ i.e ,  c₁ p²= c₁c₂² + c₁ p²

⇒ ∠ c₁c₂p = 90 °

⇒ c₂ lies on the common chord PQ and it bisects PQ.

→ since c₁c₂ ⊥ PQ and and line segments joining the centres of two circles bisects the common chord. Therefore ,c₂ is the mid  - point  PQ.

→ Hence , PQ = 2 C₂ P = 2 × 3 cm = 6 cm.

Answer 2

Answer:

Length of common chord AB is 6cm.

Step-by-step explanation:

Let the common chord be AB and P and Q be the centers of the two circles.

∴AP=5cm and AQ=3cm.

PQ=4cm        ....given

Now, seg PQ⊥ chord AB 

∴

$AR=RB= \frac{1}{2} AB$

 Cause,  perpendicular from center to the chord, bisects the chord.

Let,

PR = x cm, so, RQ = (4−x) cm

In △ARP,

${AP}^{2} = {AR}^{2} + {PR}^{2} \\ = > {AR}^{2} = {5}^{2} - {x}^{2} \: \: \: \: - - - (1)$

In △ARQ,

${AQ}^{2} = {AR}^{2} + {QR}^{2} \\ = > {AR}^{2} = {3}^{2} - ( {4 - x)}^{2} \: \: \: - - - (2)$

∴ from (1) & (2) =>

${5}^{2} - {x}^{2} = {3}^{2} - ( {4 - x)}^{2} \\ = > 25 - {x}^{2} = - 7 + 8x - {x}^{2} \\ = > 8x = 32 \\ = > x = 4 \\ \\ \sf \: substitute \: \: x = 4 \: in \: eqn.(1) \\ \\ {AR}^{2} = 25 - 16 = 9 \\ = > AR = 3 \\ \\ \therefore \: \: AB = 2 \times AR = 2 \times 3 = 6 \: cm$

So, length of common chord AB is 6cm.