Question

Не выполняя построения найдите координаты точек пересечения окружности x^2+y^2=5 и прямой x+3y=7

Answer 1

$\left \{ {{x^2+y^2=5} \atop {x+3y=7}} \right. \\\\ x=7-3y \\ \\ (7-3y)^2+y^2=5 \\ 9y^2-42y+49+y^2=5 \\ 10y^2-42y+44=0 \\ 5y^2-21+22=0 \\ \\D=441-4*5*22=441-440=1 \\ \\ y_1= \frac{21+1}{10}=2,2 \\ y_2= \frac{21-1}{10}=2$
$x_1+3*2,2=7 \\ x_1=7-6,6\\x_1=0,4 \\ \\ x_2+3*2=7 \\ x_2=7-6 \\ x_2=1$

Ответ: (2,2; 0,4) (2; 1)

Answer 2

The user's message is a request to find the coordinates of the points of intersection of the circle x^2+y^2=5 and the line x+3y=7 without drawing them. This is a mathematical problem that can be solved by combining the equations of the circle and the line in a system and solving it. I found some web search results that explain how to do this in detail . Here is a summary of the solution:

- The system of equations is:

x^2+y^2=5

x+3y=7

- To solve it, we can substitute x from the second equation into the first one and get:

(7-3y)^2+y^2=5

- Expanding and simplifying, we get:

10y^2-42y+44=0

- Solving this quadratic equation, we get:

y1=2

y2=11/5

- To find the corresponding x values, we can plug in y into the second equation and get:

x1=1

x2=-8/5

- Therefore, the coordinates of the points of intersection are:

(1,2) and (-8/5,11/5)