Two circles of radii 10cm and …

Ans.

Let X and Y be the centers of circles having radius AX = 10 cm and AY = 8 cm respectively.

Length of the chord AB = 12 cm

Suppose these circles intersect in A and B. XY intersect AB in P.

\(In \space \triangle XAY \space and \space \triangle XBY,\)

AX = BX (Radius of the circle having centre X)

XY = XY (Common)

AY = BY (Radius of the circle having centre Y)

\(\triangle XAY \cong \triangle XBY \space .......[ SSS \space criterion]\)

=> ∠AXY = ∠BXY (CPCT)

\(In \triangle XAP \space and \space \triangle XBP,\)

AX = BX (Radius of the circle)

∠AXP = ∠BXP (∠AXY = ∠BXY)

XP = XP (Common)

ΔXAP  ΔXBP (SAS congruence axiom)

=>  ∠APX = ∠BPX (CPCT)

∠APX + ∠BPX = 180° (Linear pair)

2∠APX = 180° (∠APX = ∠BPX)

=> ∠APX = 90°

=>  XP ⊥ AB

=> P is the midpoint of AB (Perpendicular from the centre to the chord, bisect the chord)

=> AP = PB = 6 cm

Let XP = x and PY = y.

In triangle AXP,

AX² = AP² + XP²

100 = 36 + x²

x²  = 64

x = 8.

XP = 8 cm.  ………..(i)

In triangle APY,

AY² = AP² + PY²

y² = 64 – 36

x²  = 28

x = √28 = 2√7.

PY = 2√7 cm.

Therefore, distance between the center of the circles = XP + PY = (8 + 2√7) cm