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 ]


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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