Question

2. Точка О – центр окружности радиуса 2. На продолжении радиуса взята точка А. Через точку А проведена касательная к окружности, где точка К – точка качания. Известно, что угол ОАК равен 60°. Найдите радиус окружности, вписанной в угол ОАК и касающейся данной внешним образом.3. Из точки расположенной вне окружности на расстоянии от центра проведена секущая, внутренняя часть которой в два раза меньше внешней и равна радиусу окружности. Найдите радиус окружности.4. Через точку М проведены две прямые. Одна из них касается некоторой окружности в точке А, другая пересекает эту окружность в точках В и С. ВС = 7, ВМ = 9. Найдите АМ.5. Дана окружность радиуса 2 с центром в точке О. Хорда АВ пересекает радиус ОС в точке Н, причем СНА равен 120°. Найдите радиус окружности, вписанной в угол АНС, и касающейся дуги АС, если ОН = . РЕШИТЕ ПОЖАЛУЙСТА ХОТЬ ЧТО НИБУДЬ, ОЧЕНЬ НАДО!!

Answer 1

Свойства касательных:
Касательная к окружности перпендикулярна к радиусу, проведенному в точку касания. 
Отрезки касательных к окружности, проведенных из одной точки, равны и составляют равные углы с прямой, проходящей через эту точку и центр окружности. 
Следовательно:
треугольники АВО и АСО прямоугольные и равные
ВО=ОС=R
ВО=АО*Sin(ВАО)
4,5 = 9*Sin(ВАО)
Sin(ВАО) = 1/2, а это синус угла 30*
Следовательно угол ВАО=САО=30*
Угол ВАС - угол между касательными
угол ВАС=угол ВАО+угол САО=60*
Ответ: угол между касательными равен 60*

 

Answer 2

Task 2: Finding the radius of the inscribed circle in angle OAK

To find the radius of the inscribed circle in angle OAK, we can use the fact that the angle OAK is 60°.

Let's denote the radius of the inscribed circle as r.

Since the inscribed circle is tangent to the line AK, we can draw a perpendicular from the center of the circle to the line AK. Let's call the point where the perpendicular intersects AK as M.

Now, we have a right triangle OAM, where OA is the radius of the given circle (which is 2) and AM is the radius of the inscribed circle (which is r).

Using trigonometry, we can find the value of r. In a right triangle, the tangent of an angle is equal to the ratio of the length of the opposite side to the length of the adjacent side.

In this case, the tangent of angle OAM is equal to the ratio of AM to OA. Since angle OAM is 90°, the tangent of angle OAM is equal to the tangent of angle OAK.

Therefore, we have the equation:

tan(60°) = r / 2

Simplifying this equation, we can solve for r:

r = 2 * tan(60°)

Using a calculator, we can find the value of tan(60°) to be approximately 1.732.

Therefore, the radius of the inscribed circle in angle OAK is:

r ≈ 2 * 1.732 ≈ 3.464

Task 3: Finding the radius of the circle

To find the radius of the circle, we can use the given information that the internal part of the secant is twice as long as the external part, and both parts are equal to the radius of the circle.

Let's denote the radius of the circle as r.

According to the given information, the internal part of the secant is twice as long as the external part, which means the external part is equal to r.

Let's denote the external part of the secant as x.

Therefore, the internal part of the secant is 2x.

The sum of the internal and external parts of the secant is equal to the length of the secant, which is equal to twice the radius of the circle:

x + 2x = 2r

Simplifying this equation, we can solve for r:

3x = 2r

x = (2/3)r

Since x is equal to the radius of the circle, we can substitute x with r:

r = (2/3)r

Simplifying this equation, we find that:

r = 0

This means that the radius of the circle is 0. However, this is not possible, as a circle cannot have a radius of 0. Therefore, there might be an error in the given information or the problem statement.

Task 4: Finding AM

To find AM, we can use the given information that one of the lines is tangent to the circle at point A, and the other line intersects the circle at points B and C.

Let's denote the length of AM as x.

Since line BC intersects the circle at points B and C, we can use the intersecting chord theorem, which states that the product of the lengths of the segments of a chord intersecting inside a circle is equal.

Therefore, we have:

BM * MC = AM * CM

Given that BC = 7 and BM = MC = 9, we can substitute these values into the equation:

9 * 9 = x * (x + 7)

Simplifying this equation, we get:

81 = x^2 + 7x

Rearranging the equation, we have:

x^2 + 7x - 81 = 0

We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = 7, and c = -81. Substituting these values into the quadratic formula, we get:

x = (-7 ± √(7^2 - 4 * 1 * -81)) / (2 * 1)

Simplifying further, we have:

x = (-7 ± √(49 + 324)) / 2

x = (-7 ± √(373)) / 2

Since the length of AM cannot be negative, we take the positive square root:

x = (-7 + √(373)) / 2

Using a calculator, we can find the approximate value of x to be:

x ≈ 4.82

Therefore, the length of AM is approximately 4.82.

Task 5: Finding the radius of the inscribed circle in angle ANS and tangent to arc AC

To find the radius of the inscribed circle in angle ANS and tangent to arc AC, we can use the given information that angle SNA is 120° and ON is equal to the radius of the given circle.

Let's denote the radius of the inscribed circle as r.

Since the inscribed circle is tangent to arc AC, we can draw a perpendicular from the center of the circle to the chord AC. Let's call the point where the perpendicular intersects AC as P.

Now, we have a right triangle ONP, where ON is the radius of the given circle (which is 2) and NP is the radius of the inscribed circle (which is r).

Using trigonometry, we can find the value of r. In a right triangle, the sine of an angle is equal to the ratio of the length of the opposite side to the length of the hypotenuse.

In this case, the sine of angle ONP is equal to the ratio of NP to ON. Since angle ONP is 90°, the sine of angle ONP is equal to the sine of angle SNA.

Therefore, we have the equation:

sin(120°) = r / 2

Simplifying this equation, we can solve for r:

r = 2 * sin(120°)

Using a calculator, we can find the value of sin(120°) to be approximately 0.866.

Therefore, the radius of the inscribed circle in angle ANS and tangent to arc AC is:

r ≈ 2 * 0.866 ≈ 1.732

Please note that the solutions provided are based on the information given in the problem statement. If there are any errors or missing information, the solutions may vary.