Question

[From (i)]
T/
I.
In the given figure. line AB is tangen
circles touching at A and B.
OA-29, BP = 18, OP - 61. then find AB.
is tangent to both the circles touching at A and B

​

Answer 1

Answer:

OP = 61 units

Step 1:

Referring to the figure attached below, we will draw a perpendicular line from point P to the radius OA of the bigger circle at M, i.e.,

∠AMP = 90° …. (i)

We know that the tangent to a circle is perpendicular to the radius at the point of tangency.

∴ ∠OAB = 90° and ∠PBA = 90° …. (ii)

Consider quadrilateral AMPB, applying angle sum property, we get

∠OAB + ∠PBA + ∠AMP + ∠BPM = 360°

⇒ 90° + 90° + 90° + ∠BPM = 360°

⇒ ∠BPM = 360° - 270°

⇒ ∠BPM = 90° …… (iii)

Step 2:

From (i), (ii) & (iii), we get

Quadrilateral AMPB is a rectangle

∴ AM = BP = 18 units …… [opposite sides of a rectangle are equal in length]

∴ OM = OA – AM = 29 – 18 = 11 units ….. (iv)

Now, in right-angled ∆ PMO, applying Pythagoras theorem,

OP² = OM² + PM²

⇒ 61² = 11² + PM² ….. [OP = 61 (given) and OM = 11 (from (iv))]

⇒ PM² = 3721 – 121

⇒ PM = √3600

⇒ PM = 60 units

∴ PM = AB = 60 ← opposite sides of rectangle AMPB are equal in length

Answer 2

Answer:

$\huge\star\mathfrak\blue{{Answer:-}}$

OP = 61 units

Step 1:

Referring to the figure attached below, we will draw a perpendicular line from point P to the radius OA of the bigger circle at M, i.e.,

∠AMP = 90° …. (i)

We know that the tangent to a circle is perpendicular to the radius at the point of tangency.

∴ ∠OAB = 90° and ∠PBA = 90° …. (ii)

Consider quadrilateral AMPB, applying angle sum property, we get

∠OAB + ∠PBA + ∠AMP + ∠BPM = 360°

⇒ 90° + 90° + 90° + ∠BPM = 360°

⇒ ∠BPM = 360° - 270°

⇒ ∠BPM = 90° …… (iii)

Step 2:

From (i), (ii) & (iii), we get

Quadrilateral AMPB is a rectangle

∴ AM = BP = 18 units …… [opposite sides of a rectangle are equal in length]

∴ OM = OA – AM = 29 – 18 = 11 units ….. (iv)

Now, in right-angled ∆ PMO, applying Pythagoras theorem,

OP² = OM² + PM²

⇒ 61² = 11² + PM² ….. [OP = 61 (given) and OM = 11 (from (iv))]

⇒ PM² = 3721 – 121

⇒ PM = √3600

⇒ PM = 60 units

∴ PM = AB = 60 ← opposite sides of rectangle AMPB are equal in length