Two circles with radii 5cm and 8 cm touch each other externally at a point A. If a straight line through t h e point Acuts the circles at points P and Q respectively, the AP : AQ isa)8 : 5b)5 : 8c)3 : 4d)4 : 5Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Solution:

Given, two circles with radii 5 cm and 8 cm touch each other externally at point A.

Let O1 and O2 be the centres of the circles with radii 5cm and 8 cm respectively.

Join O1A and O2A.

Let the straight line through A intersect the circles at points P and Q respectively.

Join OP and OQ.

We need to find AP : AQ.

Let AP = x and AQ = y.

To find x and y, we need to use the property that tangents drawn to a circle from an external point are equal in length.

Hence, PA = AQ.

By Pythagoras theorem, in right triangle O1AP, we have

O1P^2 = O1A^2 + AP^2

=> O1P^2 = (5)^2 + x^2

=> O1P^2 = 25 + x^2

Similarly, in right triangle O2AQ, we have

O2Q^2 = O2A^2 + AQ^2

=> O2Q^2 = (8)^2 + y^2

=> O2Q^2 = 64 + y^2

Since PA = AQ, we have

x + y = O1P + O2Q

=> x + y = OA

=> x + y = 13

Squaring both sides, we get

x^2 + 2xy + y^2 = 169

Substituting the values of O1P^2 and O2Q^2, we get

25 + x^2 + 2xy + 64 + y^2 = 169

=> x^2 + y^2 + 2xy = 80

=> (x + y)^2 - 2xy = 80

=> 13^2 - 2xy = 80

=> 169 - 80 = 2xy

=> 2xy = 89

=> xy = 89/2

Therefore, AP : AQ = x : y = 5 : 8.

Hence, the correct option is (b) 5 : 8.

Answer

c) 3 : 4

If two circles with radii 5cm and 8cm touch each other externally at a point A, the distance between the centers of the circles will be the sum of the radii, or 5 + 8 = 13cm.

If a straight line through the point A cuts the circles at points P and Q, the distances from the centers of the circles to points P and Q will be 3cm and 4cm, respectively.

Thus, the ratio of AP to AQ will be 3 : 4, so the correct answer is c) 3 : 4.

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