Question

AB = 18 cm, CD = 7 cm are two chords of a circles with centre O and they
extended in the same direction meet at P. From P, a tangent PT is drawn to to
the circle at P and PB = 6 cm, then
(A) PT = 12 cm (B) PA = 24 cm (C) PO = 16 cm (D) PC = 16 cm​

Answer 1

Answer:

We join OA and drop perpendiculars OM & ON from O to CD & AB respectively.

So CM=

2

1

CD & AN=

2

1

BD.........(i) (since the perpendicular

from the centre of a circle to its chord bisects the lattar).

Also AB=AP+PB=(4+6) cm=10cm.

Now the chords AB & CD intersect at P within the given circle.

∴ PC× PD=AP× BP

⟹ PD=

PC

AP×BP

=

2

4×6

cm=12 cm.

∴ CD=PD+PC=(12+2) cm=14 cm.

∴ CM =

2

1

×14cm=7 cm and

AN=

2

1

×10 cm =5 cm (from i).

So PM=CM−PC=(7−2) cm=5 cm.

Now the quadrilateral OMPN has three of its angles =90

o

each.

∴ The quadrilateral OMPN is a rectangle.

i.e ON=PM=5 cm.

Let us consider the triangle OAN.

∠ ANO=90

o

, ON=5 cm and AN=5 cm.

∴Δ OAN is a right one with OA, which is the radius$$=r

of the circle, as its hypotenuse.

So, by Pythagoras theorem, we get OA

2

=(ON

2

+AN

2

)=(5

2

+5

2

)cm=50 cm=r

2

.

So ar.(circle)=πr

2

=π×50cm

2

=50πcm

2

.

Ans- Option C.