Question

[Maths]
Topic: Circles

Q) In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.
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Diagram in attachment.
No spams, please!​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

At first, consider the triangle on the left.

→ AM is perpendicular to PM as PM is the tangent.

Apply Pythagoras Theorem on ∆AMP:

→ AP² = AM² + PM²

→ 17² = AM² + 15²

→ 289 = AM² + 225

→ AM² = 289 - 225

→ AM² = 64

→ AM = 8 cm.

Now, notice that AM is the radius of the circle.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Now, consider the second triangle.

BN is perpendicular to QN as QN is the tangent.

Applying Pythagoras Theorem on ∆BQN:

→ BQ² = QN² + BN²

→ 13² = 12² + BN²

→ BN² = 13² - 12²

→ BN² = 5²

→ BN = 5 cm.

Notice that BN is also the radius of the circle.

Therefore, the distance between A and B will be:

→ AB = AM + BN

→ AB = 8 + 5 cm

→ AB = 13 cm.

★ Which is our required answer.

$\textsf{\large{\underline{Answer}:}}$

• The distance between the centres A and B is 13 cm.

Answer 2

Topic:-

• Circles.

Question:-

In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

Given:-

• Two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm.

To Find:-

• The distance between the centres A and B of the circles.

Solution:-

AM is radius and PM is tangent.

∴ AM ⊥ PM.

Similarly, BN ⊥ NQ.

Now in right ∆APM,

⇒ AP² = AM² + PM²

⇒ 17² = AM² + 15²

⇒ AM² = 17² – 15²

⇒ AM² = 289 – 225 = 64 = (8)²

∴ AM = 8 cm.

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Similarly, in right ∆BNQ.

⇒ BQ² = BN² + NQ²

⇒ 13² = BN² + 12²

⇒ 169 = BN² + 144

⇒ BN² = 169 – 144 = 25 = (5)²

∴BN = 5 cm.

_________________________________________

Now AB = AM + BN

(AR = AM and BR = BN).

⇒ AB = 8 + 5

⇒ AB = 13 cm.

Answer:-

${ \boxed{ \sf \large \color{red} \therefore The \: distance \: between \: the \: centres \: A \: and \: B \: of \: the \: circles = 13 \: cm.}}$

Hope you have satisfied. ⚘