# Thread: question on properties of circle 8

1. ## question on properties of circle 8

Referring to the attached pic:
APQ is a semi-circle with PQ as the diameter.
QP is extended to B, and BA is a tangent to the circle at A.
AC is perpendicular to PQ and PC:CQ = 1:4.
AB = 8cm
(a) Prove $\displaystyle AC^2 = (PC)(QC)$
(b) Prove $\displaystyle AB^2 = (BP)(BQ)$
(c) If BP = y and PC = k, prove that $\displaystyle k = \frac{3}{5} y$
(d) Find BP
(e) Find the radius of circle.

My work:
(a)
triangles APC ~ ACQ (same angles)
hence AC:PC = QC:AC
$\displaystyle AC^2 = (PC)(QC)$
(b)
triangles BAQ ~ BPA (same angles)
hence AB:BQ = BP:AB
$\displaystyle AB^2 = (BP)(BQ)$
(c)
since PC:CQ = 1:4,
CQ = 4k and hence PQ = 5k
hence $\displaystyle \frac{k}{AC} = \frac{AC}{4k}$
$\displaystyle AC^2 = 4k^2 = (2k)^2$
hence we have AC = 2k
$\displaystyle AP^2 = AC^2 + PC^2$ (Pythagarus)
$\displaystyle AP^2 = (2k)^2 + k^2$
$\displaystyle AP^2 = 5k^2$
AP = (k)(5^0.5)
Also $\displaystyle AQ^2 = PQ^2 - AP^2$
$\displaystyle AQ^2 = (5k)^2 - 5k^2$
$\displaystyle AQ^2 = 20k^2$
AQ = (2k)(5^0.5)
y:8 = AP:AQ = k(5^0.5):2k(5^0.5)
y:8 = 1:2
hence y = 4
and 8:BQ = AP:AQ = 1:2
8:(5k + y) = 1:2
8:(5k + 4) = 1:2
16 = 5k + 4
k = $\displaystyle \frac{12}{5}$
Therefore, $\displaystyle \frac{3}{5} y = (\frac{3}{5})(4) = \frac{12}{5} = k$
Confused here: relation between k and y is supposed to be proved without working out their values in advance. Is there another method to prove that k = 3y/5 ????
(d)
BP = y = 4cm
(e)
PQ = 5k = 12cm

2. From circle 1 to circle 8 ... learning the hula hoop ?