In Fig. 9.15, from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. Prove that
(i) PA . PB = PN² - AN²
(ii) PN² - AN² = OP² - OT²
(iii) PA.PB = PT²
Solution:
Given, PT is a tangent to a circle from an external point P.
A line segment PAB is drawn to a circle with centre O.
ON is perpendicular on the chord AB.
(i) We have to prove that PA . PB = PN² - AN²
Considering LHS: PA . PB
From the figure,
PA = PN - AN
PB = PN + BN
So, PA . PB = (PN - AN)(PN + BN)
Since, ON is perpendicular on the chord AB.
AN = BN
Now, PA . PB = (PN - AN)(PN + AN)
By using algebraic identity,
(a² - b²) = (a + b)(a - b)
So, (PN - AN)(PN + AN) = PN² - AN²
= RHS
Therefore, PA . PB = PN² - AN² -------------- (a)
(ii) We have to prove that PN² - AN² = OP² - OT²
Considering LHS: PN² - AN²
Since ON ⟂ PN
PN² - AN² = (OP² - ON²) - AN²
= OP² - (ON² + AN²)
From the figure,
ON² + AN² = OA²
PN² - AN² = OP² - OA²
Now, OA = OT = radius
So, OP² - OA² = OP² - OT² = RHS
Therefore, PN² - AN² = OP² - OT² -------------- (b)
(iii) WE have to prove that PA.PB = PT²
From (a) and (b),
PA . PB = OP² - OT²
From the figure,
In triangle OPT,
OP² - OT² = PT²
Therefore, PA . PB = PT²
✦ Try This: PT is a tangent and PAB is a secant. If PT = 6 cm and AB = 5 cm, then find the length of PA.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.4 Sample Problem 1
In Fig. 9.15, from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. Prove that (i) PA . PB = PN² - AN², (ii) PN² - AN² = OP² - OT², (iii) PA.PB = PT²
Summary:
In Fig. 9.15, from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. It is proven that (i) PA . PB = PN² - AN², (ii) PN² - AN² = OP² - OT², (iii) PA.PB = PT²
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