Question

In the below figure PT is tangent to a circle with
centre 0, PT = 12 cm, AP= 8 cm. Find the
radius of the circle,

plz help​

Answer 1

Given :

• PT is tangent to a circle with centre O.
• PT = 12 cm.
• AP= 8 cm.

To Find :

The radius of the circle.

Solution :

Let's understand first!!

Here, PT is the tangent to the circle with centre O.

So, OA = OT (radii of the circle)

We are given the AP.

∵ OP = OA + AP

We know that the tangent is perpendicular to the radius which joins the centre of the circle to the point at the circumference.

Here, OT ⊥ PT (above mentioned theorem)

• ∠OTP = 90°

∴ ∆PTQ is a right angled triangle.

Let the radius of the circle be “x” cm.

• OP = OA + AP = (x + 8) cm
• OT = x cm
• PT = 12 cm

By using Pythagoras theorem,

⇒ (Hypo)² = (Base)² + (Side)²

⇒ (OP)² = (OT)² + (PT)²

Substituting the values,

⇒ (x + 8)² = (x)² + (12)²

⇒ x² + 2.x.8 + 64 = x² + 144

⇒ x² + 16x + 64 = x² + 144

Cancelling x² from both sides,

⇒ 16x + 64 = 144

⇒ 16x = 144 - 64

⇒ 16x = 80

⇒ x = 80/16

⇒ x = 5

∴ x = 5.

• OT = x = 5 cm.

The radius of the circle is 5 cm.

Answer 2

$\sf\underline\color{Black}Given$

PT=30cm

OT is the radius of a circle

$\tt\therefore OT \:will \:be\: half \:of \:given\: diameter\:OT = 16cm$

∠OTP = 90⁰

• (tangent is perpendicular to radius at point of contact)

$\tt\color{purple}\underline{Using \: Pythagoras \: theorem}$

$→\sf★(OP)^2 =(OT)^2 +(TP)^2★$

$→\tt{OP^{2} = {16}^{2} + {30}^{2} }$

$→\tt{OP^{2} = 1156 }$

$→\tt{OP = \sqrt{1156} }$

• $\sf \large { \underline{ \underline{\fbox {\red{OP = 34cm}}}}}$

OP = 34cm