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Two circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
- a)30°
- b)45°
- c)60°
- d)90°
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Two circles touch each other externally at C and AB is a common tangen...
Here OA ⊥ AB and O'B ⊥ AB, then ∠OAB = ∠O'BA = 90° And ∠AOO' = ∠BO'O = 90°
∴ ABO'O is a rectangle.
Also, OC = OA, then ∠OCA = ∠OAC= x [Let] ∴ 90° (x + x) = 180°
⇒ x = 45° Again, O'C = O'B, then ∠OC'B = ∠O'BC = y [Let] ∴ 90° (y + y) = 180°
⇒ y = 45°
Now, ∠OCA + ∠ACB + ∠BCO' = 180° [Straight angle]
⇒ 45° + ∠ACB + 45° = 180°
⇒ ∠ACB = 90°
∴ ABO'O is a rectangle.
Also, OC = OA, then ∠OCA = ∠OAC= x [Let] ∴ 90° (x + x) = 180°
⇒ x = 45° Again, O'C = O'B, then ∠OC'B = ∠O'BC = y [Let] ∴ 90° (y + y) = 180°
⇒ y = 45°
Now, ∠OCA + ∠ACB + ∠BCO' = 180° [Straight angle]
⇒ 45° + ∠ACB + 45° = 180°
⇒ ∠ACB = 90°
Most Upvoted Answer
Two circles touch each other externally at C and AB is a common tangen...
Since AB is a common tangent to the circles, it will be perpendicular to the line joining the centers of the circles at the point of contact (C).
Let O1 and O2 be the centers of the circles, and let r1 and r2 be their respective radii.
Since AB is perpendicular to the line joining the centers of the circles, triangle O1CO2 is a right triangle.
Let OC = d, which is the distance between the centers of the circles.
Using the Pythagorean theorem in triangle O1CO2, we have:
d^2 = (r1 + r2)^2
Now, let's consider triangle O1AC. It is a right triangle with hypotenuse O1C and leg AC.
By the Pythagorean theorem in triangle O1AC, we have:
AC^2 = O1C^2 - O1A^2
AC^2 = d^2 - r1^2
Similarly, in triangle O2BC, we have:
BC^2 = O2C^2 - O2B^2
BC^2 = d^2 - r2^2
Since AB is a common tangent to the circles, AC = BC.
Therefore, we can equate the expressions for AC^2 and BC^2:
d^2 - r1^2 = d^2 - r2^2
Simplifying this equation, we have:
r1^2 = r2^2
So, the radii of the two circles are equal.
Let O1 and O2 be the centers of the circles, and let r1 and r2 be their respective radii.
Since AB is perpendicular to the line joining the centers of the circles, triangle O1CO2 is a right triangle.
Let OC = d, which is the distance between the centers of the circles.
Using the Pythagorean theorem in triangle O1CO2, we have:
d^2 = (r1 + r2)^2
Now, let's consider triangle O1AC. It is a right triangle with hypotenuse O1C and leg AC.
By the Pythagorean theorem in triangle O1AC, we have:
AC^2 = O1C^2 - O1A^2
AC^2 = d^2 - r1^2
Similarly, in triangle O2BC, we have:
BC^2 = O2C^2 - O2B^2
BC^2 = d^2 - r2^2
Since AB is a common tangent to the circles, AC = BC.
Therefore, we can equate the expressions for AC^2 and BC^2:
d^2 - r1^2 = d^2 - r2^2
Simplifying this equation, we have:
r1^2 = r2^2
So, the radii of the two circles are equal.
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Two circles touch each other externally at C and AB is a common tangent to the circles. Then,∠ACB =a)30°b)45°c)60°d)90°Correct answer is option 'D'. Can you explain this answer?
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