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 

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