Question

As shown in the
figure, AP is tangent
to the circle at point
A. Secant through P
intersects chord AY
in point X, such that Z
AP=PX = XY.
If PQ = 1 and QZ = 8;
then find AX.
4. In the figure, M is the point of contact of two
internally touching circles. The chord CD of the
bigger circle touches the smaller circle at point N.
Line AMB is their common tangent. Prove MN
bisects ZCMD.
(HOTS).​

Answer 1

Answer:

PQ × PZ = PA²

=) PQ[PQ + QZ] = PA²

=) 1(1+8)= PA²

=) 1(9)= PA²

=) PA² = 9

=) PA² = 3²

=) PA= 3cm

So,

PA = PX = XY = 3cm

Now,

PX = 3cm

=) PQ + QX= 3

=) 1 + QX = 3

=) QX= 3-1

=) QX= 2cm

Now,

QZ = 8cm

=) ZX + QX = 8

=) ZX + 2 = 8

=) ZX = 8-2

=) ZX = 6cm

We know that when two chords of a circle intersect internally, then product of the length of segments are equal.

So,

AX × XY = ZX × QX

=) AX × 3 = 6 × 2

=) 3AX = 12

=) AX= 12/3

=) AX = 4cm

Thus,

AX is the required of 4cm.

OK or Question hai to btao