Question

hello plx explain this answer ...and correct full explaination ​

Answer 1

We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.

∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ

Now, in right ∆OLP, we have

PL² + OL² = 2

⇒ PL² + (4 – x)2 = 52

⇒ PL² = 52 – (4 – x)2

⇒ PL² = 25 -16 – x2 + 8x

⇒ PL² = 9 – x2 + 8x …(i)

Again, in right ∆O’LP,

PL² = PO‘² – LO‘²

= 32 – x² = 9 – x² …(ii)

From (i) and (ii), we have

9 – x² + 8x = 9 – x²

⇒ 8x = 0

⇒ x = 0

⇒ L and O’ coincide.

∴ PQ is a diameter of the smaller circle.

⇒ PL = 3 cm

But PL = LQ

∴ LQ = 3 cm

∴ PQ = PL + LQ = 3cm + 3cm = 6cm

Thus, the required length of the common chord = 6 cm.

Therefore

The length of the common chord = 6cm.

Answer 2

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