Q. 14.6( 20 Votes )

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Answer :

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

Radius of circle C1, r1 = 4 cm

Radius of circle C2, r2 = 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 ]


(OB)2 + (BC)2 = (OC)2

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


(4)2 + (BC)2 = (5)2

16 + (BC)2 = 25

(BC)2 = 25 - 16 = 9

BC = 3 cm


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


AC = AB + BC

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

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