F4 math problems ( Circles )

( 1 ) By properties of tangents,

AP = PR = 2cm , BQ = QR = 3cm and BT = AT

Hence PQ = 5 cm

Let PT = x and QT = y.

Considering ΔPQT,

x2 + y2 = 52 ( Pyth. Theorem ) --- ( 1 )

3 + y = 2 + x ( since BT = AT )

x = y + 1 --- ( 2 )

Put ( 1 ) into ( 2 ).

( y + 1 )2 + y2 = 25

y2 + 2y + 1 + y2 = 25

y2 + y – 12 = 0

( y – 3 )( y + 4 ) = 0

y = 3 or y = - 4 ( rejected )

So x = 3 + 1 = 4

Then BT = AT = 6cm

Besides, TA ⊥ TB and the tangents are ⊥ radius with OB = OA, therefore OBTA is a square.

Hence the radius of the circle is 6cm.


( 2 ) PA = PB ( properties of tangents )

To proveΔAPB ~ΔRPQ

∠APB = ∠RPQ ( common ∠ )

∠PAB = ∠PRQ ( corr. ∠s , AB // RQ )

SoΔAPB ~ΔRPQ ( AA )

Hence, PQ = QR ( corr. sides, ~Δ )

1.
Let r be the radius of the circle
AP = PR = 2 [reason?]
BQ = QR = 3 [reason?]
PQ = PR + RQ = AP + BQ = 2 + 3 = 5
OA = OB = AT = BT = r [reason?]
PT^2 + QT^2 = PQ^2 (path. th.)
=> (r - AP)^2 + (r - BQ)^2 = 5^2
=> (r - 2)^2 + (r - 3)^2 = 25
=> 2r^2 -10r -12 = 0
=> r^2 - 5r -6 = 0
=> (r - 6)(r + 1) = 0
=> r = 6 (r = -1 rejected)
So the radius of the circle is 6

2.
PA = PB [reason ?]
PA/PR = PB/PQ (since AB // RQ)
=> PQ = PR (since PA = PB)