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Two circles toouch each other externally at C and AB is the common tangent to the circles. Then angle ACB is: (a) 60 degrees (b) 45 degrees (c) 30 degrees (d) 90 degrees

Two circles toouch each other externally at C and AB is the common tangent to the circles. Then angle ACB is:
(a) 60 degrees
(b) 45 degrees
(c) 30 degrees
(d) 90 degrees 

Grade:10

4 Answers

ruchi yadav
askIITians Faculty 27 Points
7 years ago
If circles are of same radius then ACB =90 deg.


Thank You,
Ruchi
Askiitians Faculty
Yash Kandpal
37 Points
7 years ago
@ Ruchi Yadav But the radii are not specified
Nithin
36 Points
7 years ago
Even if the radius is not same it will be 90 deg. There is need for the figure in this question. I have the fig at (https://drive.google.com/file/d/0B2EDcEci0le6UEdzQjdCcFlTcUU/edit?usp=sharing) Here`s the explanation: Let O be the center of larger circle and O1 be the center of the smaller circle. Now join AO and CO(both are the radii of larger circle). Then join BO1 and CO1(both are the radii of smaller circle). You will notice that : ang.OAC = ang.OCA similarly, ang.O1BC = ang.O1CB Now in triangle ACB ; ang.ACB = 180 - (ang. CAB + ang. CBA) ---------------- (1) But, ang .CAB = ang . OAB - ang. OAC =90 - ang . OAC =90 - ang . ACO ----------------------------------------- (2) Similarly, ang. CBA = 90 - ang. BCO1-------------------------------- (3) Substituting (2) and (3) in (1) we get, ang. ACB = 180 - (90 - ang. ACO + 90 - ang. BCO1) ang. ACB = 180 - (180 - ang. ACO - ang. BCO1) ang. ACB = 180 - 180 + ang. ACO + ang. BCO1 ang. ACB = ang. ACO + ang. BCO1 ---------------------- (4) But, OCO1 is a straight line i.e; ang. ACO + ang. ACB + ang. BCO1 = 180 ang. ACB = 180 - (ang. ACO + ang. BCO1) ------------ (5) Substituting (4) to (5) We get, ang. ACB = 180 - ang. ACB ang. ACB X 2 = 180 ang. ACB = 180/2 = 90 deg.---------------(ANSWER)
Yash Kandpal
37 Points
7 years ago
@ Nithin thanks a lot

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