Question

Прямая l касается окружности w в точке K. На окружности выбрана точка Р; Н- проекция Р на прямую l (отрезок РН не имеет общих точек с w кроме точки Р). Найдите радиус w, если РН=4, НК=9 Срочно!!!!! помогите пожалуйста ​

Answer 1

Ответ:

Радиус окружности ω равен $\displaystyle 12\frac{1}{8}$  ед.

Пошаговое объяснение:

Требуется найти радиус окружности ω.

Дано: ω(О, R)

l - касательная;

К - точка касания;

Н- проекция Р на прямую l;

РН = 4, НК = 9.

Найти: R

Решение:

• Проекция точки на прямую – это основание перпендикуляра, опущенного из этой точки на заданную прямую.

⇒ РН ⊥ НК

• Радиус, проведенный в точку касания перпендикулярен касательной.

⇒ ОК ⊥ НК

Проведем РМ ⊥ ОК.

Получим РНКМ - прямоугольник.

• Противоположные стороны прямоугольника равны.

⇒ НК = РМ = 9; РН = МК = 4.

Рассмотрим ΔРМО - прямоугольный.

ОР = ОК = R.

Пусть ОР = х, тогда МО = (х - 4)

По теореме Пифагора:

ОР² = МО² + РМ²

х² = (х - 4)² + 9²

х² = х² - 8х + 16 + 81

8х = 97   |:8

$\displaystyle x=\frac{97}{8}\\ \\x=12\frac{1}{8}$

Радиус окружности ω равен $\displaystyle 12\frac{1}{8}$  ед.

Answer 2

Problem Analysis

We are given a circle with a point K on it. A point P is chosen on the circle, and the projection of P onto a line l is denoted as N. The line segment PN does not intersect the circle except at point P. We need to find the radius of the circle, given that PN = 4 and NK = 9.

Solution

To solve this problem, we can use the properties of right triangles and the Pythagorean theorem. Let's denote the center of the circle as O and the radius as r.

1. Draw a line segment OK from the center O to the point K on the circle. Since OK is a radius of the circle, its length is r.

2. Draw a line segment OP from the center O to the point P on the circle. Since OP is also a radius of the circle, its length is also r.

3. Draw a line segment NP perpendicular to the line l, with N as the point of intersection between line l and line segment PN.

4. We are given that PN = 4 and NK = 9. Therefore, the length of line segment PK is PN + NK = 4 + 9 = 13.

5. Since line segment NP is perpendicular to line l, we can form a right triangle PNO, where NO is the hypotenuse, NP is the height, and OP is the base.

6. According to the Pythagorean theorem, the square of the hypotenuse (NO) is equal to the sum of the squares of the other two sides (NP and OP). Therefore, we have the equation:

NO^2 = NP^2 + OP^2

7. We can substitute the values we know into the equation. NP is equal to the length of line segment PK, which is 13, and OP is equal to the radius r. Therefore, the equation becomes:

NO^2 = 13^2 + r^2

8. We also know that NO is equal to the radius r. Therefore, we can rewrite the equation as:

r^2 = 13^2 + r^2

9. Simplifying the equation, we get:

0 = 13^2

10. This equation has no solution, which means there is no valid radius that satisfies the given conditions. Therefore, there is no solution to this problem.

Conclusion

Based on the given information, there is no valid radius for the circle that satisfies the conditions where PN = 4 and NK = 9.