AB is a diameter of circle( O,15 ) . A tangent is drawn from B to circle ( O, 9) which
touches circle ( O,9) at D. BD intersects circle( O,15 ) in C. Find AC
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2
Answer:
Given- O is the centre of a circle whose diameter is BC. AB is a chord and OD⊥ AB. BD=5cm and OD=4cm. CD has been joined.
To find out- CD=?
Solution- OD⊥AB.
∴ D is the mid point of AB since the perpendicular, dropped from the center of a circle to its any chord bisects the latter. So AB=2BD=2×5cm=10cm. And BD=AD=5cm. Now ∠BAC=90
o
since angle in a semicircle=90
o
. ∴ΔCAB&ΔCDB are right triangles with BC&DC as hypotenuses.
∴ By Pythagoras theorem, we have OB=
BD
2
+OD
2
=
5
2
+4
2
cm=
41
cm.
But BC=2OB(diameter=2radius).
BC=2×
41
cm.
∴AC=
BC
2
−AB
2
=
(
41
)
2
−10
2
cm=8cm.
So CD=
AD
2
+AC
2
=
5
2
+8
2
cm=
89
cm.
Ans- Option C.
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