AB is a chord of length 24cm of a circle of radius 13cm. The tangents at A and B intersect at a pt. C. Find the length AC.
Let O be the centre of the circle and AB and OC intersect at D
Now let CD = y
Since OC is the perpendicular bisector of AB
In right ∆ OAP
OA2 = AD2 + OD2
⇒ OD2 = OA2 – AD2 = (13 cm)2 – (12 cm)2 = 169 cm2 – 144 cm2 = 25 cm2
In right ∆'s OAC and ADC
AC2 =AD2 + DC2
OC2 = OA2 + AC2
⇒ OC2 = OA2 + (AD2 + DC2)
⇒ (CD + OD)2 = (13 cm)2 + (12 cm)2 + y2
⇒ (y + 5 cm)2 = 169 cm2 + 144 cm2 + y2
⇒ y2 + 10y cm + 25 cm2 = 313 cm2 + y2
⇒ 10y = (313 – 25) cm = 288 cm
⇒ CD = 28.8 cm2
∴ AC2 = AD2 + DC2