Question

In Fig. 17.21, PT = 2 7 cm, OP = 8 cm, find the radius of the circle, if O is the centre of the circle

Answer 1

Answer:

6 CM

Step-by-step explanation:

u can use Pythagoras here

Answer 2

• The radius of the given circle is equal to 6 cm .

Correct Question :- In Fig. 17.21, PT = 2√7 cm, OP = 8 cm, find the radius of the circle, if O is the centre of the circle ?

Concept used :-

• secant - tangent theorem :- If PAB is a secant to a circle intersecting the circle at A and B, and PT is a tangent to the circle at T, then :- PA × PB = PT² .

Solution :-

Let us assume that, radius of the circle is equal to r cm .

So,

→ OB = OA = r cm { Radius }

now, given that,

→ OP = 8 cm

→ OP - OA = AP

→ (8 - r) = AP

→ PA = (8 - r) cm ------- Equation (1)

and,

→ PB = PO + OB

→ PB = (8 + r) cm ----------- Equation (2)

also, given that,

→ PT = 2√7 cm ---------- Equation (3)

putting values of Equation (1), Equation (2) and Equation (3) in secant - tangent theorem we get,

→ PA × PB = PT²

→ (8 - r)(8 + r) = (2√7)²

using (a - b)(a + b) = a² - b² in LHS,

→ (8)² - (r)² = (2)² × (√7)²

→ 64 - r² = 4 × 7

→ 64 - r² = 28

→ r² = 64 - 28

→ r² = 36

→ r² = (±6)²

square root both sides,

→ r = ± 6 cm

since radius of circle cant be in negative . Hence, radius of the circle is equal to 6 cm .

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