Multiple choice

Two circles with centres $A$ and $B$ of radii $3\ cm$ and $4\ cm$, respectively intersect at two points $C$ and $D$ such that $AC$ and $BC$ are tangents to the two circles. Find the 10 times length of the common chord $CD$

  1. 48

  2. 58

  3. 56

  4. 54
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In triangle ABC, AC=3, BC=4. Since AC and BC are tangents, they are perpendicular to the radii at the points of contact. This implies angle ACB = 90 degrees. AB = sqrt(3^2 + 4^2) = 5. The common chord CD is perpendicular to AB. Area of triangle ABC = 1/2 * 3 * 4 = 6. Also area = 1/2 * AB * h = 1/2 * 5 * h. h = 12/5. CD = 2 * h = 24/5 = 4.8. 10 * CD = 48.

AI explanation

Because AC and BC are tangents to the circles at point C, the radius is perpendicular to the tangent at the point of contact, making angle ACB equal to 90 degrees. This forms a right-angled triangle ABC with legs AC = 3 cm and BC = 4 cm, making the distance between the centers AB equal to 5 cm. The common chord CD is perpendicular to the line of centers AB, intersecting it at point M. The area of triangle ABC is 1/2 * 3 * 4 = 6, and also equals 1/2 * AB * CM, which gives CM = 12/5 cm. Since M is the midpoint of the common chord, the full length of CD is 2 * 12/5 = 24/5 cm. Multiplying this length by 10 gives 10 * 24/5 = 48. The correct option is A.

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