Answer :

Given : Two circles (say C1 and C2) with common center as O and

Radius of circle C1, r_{1} = 4 cm

Radius of circle C2, r_{2} = 5 cm

Also say AC is the chord of circle C2 which is tangent to circle C1 and OB is the radius of circle to the point of contact of tangent AC .

To find : Length of chord AC

Now, Clearly OB ⏊ AC [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

So OBC is a right-angled triangle

So, it will satisfy Pythagoras theorem [ i.e. (base)^{2} + (perpendicular)^{2} = (hypotenuse)^{2} ]

i.e.

(OB)^{2} + (BC)^{2} = (OC)^{2}

As OB and OC are the radii of circle C1 and C2 respectively.

So

(4)^{2} + (BC)^{2} = (5)^{2}

16 + (BC)^{2} = 25

(BC)^{2} = 25 - 16 = 9

BC = 3 cm

Also

AB = BC [ Perpendicular through the center to a chord in a circle (C2 in this case) bisects the chord]

And

AC = AB + BC

= AB + AB = 2AB = 2(3) = 6 cm

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