Two circles of radii 3 and 12 touch outwardly at point K. Both circles touch one straight line

Two circles of radii 3 and 12 touch outwardly at point K. Both circles touch one straight line: the large one – at point A, the smaller one – at point B. Line AK intersects the smaller circle at point C, line BK intersects the large circle at point D. Find the area quadrilateral ABCD.

Let two circles O and O₁ with radii OA = 12 and O₁B = 3 touch each other externally at point K. Both circles touch one straight line AB: the larger one at point A, the smaller one at point B. Line AK intersects the smaller circle at point C, straight line BK intersects the great circle at point D. To find the area of ​​the quadrilateral ABCD, we prove that it is a rectangular trapezoid ABCD with bases AD = 2 ∙ ОА = 24 and CB = 2 ∙ О₁В = 6. Let point М be the point of intersection of tangent lines MC and AB. Hence, MA and MK are tangents drawn from point M to the first circle, MK and MВ are tangents drawn from point M to the second circle. By the property of tangents MK = MA and MK = MВ. We get that MK = MA = MВ. Then KM is the median of the triangle AKВ, equal to half of the side to which it is drawn KM = ½ ∙ BC. Consider additionally a circle circumscribed around ∆ AKВ centered at point M, radius KM, in it ∠AKB = 90 °, since it rests on the diameter, it turns out that ∆ AKВ is rectangular. ∠ DKС = ∠AKВ = 90 °, as vertical, then the adjacent for ∠ DKС and ∠AKВ will be ∠ AKD = ∠ВKС = 90 °, and AD is the diameter of the first circle and BC – the diameter of the second circle, perpendicular to AB. So AD | | BC and quadrilateral ABCD is a rectangular trapezoid ABCD with bases AD and CB. The height of the trapezoid AB is found by the Pythagorean theorem (OK + O₁K) ² = AB² + (AO – O₁B) ² or 15² = AB² + 9²; AB = 12. Area of ​​the trapezoid S (ABCD) = (AD + BC) ∙ AB / 2 = (24 + 6) ∙ 12/2 = 180.
Answer: The area of ​​the trapezoid is 180 square units.