Trig Question?

Question

The Line 3y+X=10 intersetcs the circle x^2 + y^2 = 50 in two distinct points

Find the coordinates of the two points of intersection

Answer 1

Solve the linear equation for one variable...

3y + x = 10
x = 10 - 3y

Now put that into the circle equation and solve for y.

x^2 + y^2 = 50
(10-3y)^2 + y^2 = 50
100 - 60y +9y^2 + y^2 = 50
10y^2 - 60y + 100 = 50
10(y^2 - 6y + 10) = 50
y^2 - 6y + 10 = 5
y^2 - 6y + 5 = 0
(y - 5)(y - 1) = 0
y = 5, 1

Now put those back into the linear equation to get x.

y=5
x = 10 - 3(5) = -5

y = 1
x = 10 - 3(1) = 7

(-5, 5) and (7, 1) are the points of intersection

Answer 2

3y + x = 10, x^2 + y^2 = 50

Your first step is to solve for x in the linear equation.

x = 10 - 3y

Now, plug this value of x into the second equation (the circle);

(10 - 3y)^2 + y^2 = 50
100 - 60y + 9y^2 + y^2 = 50

Moving everything to the left hand side, and reordering everything in descending power of y,

10y^2 - 60y + 50 = 0

Dividing everything by 10,

y^2 - 6y + 5 = 0

This factors as

(y - 5)(y - 1) = 0
Giving us the solutions

y = {1, 5}

To find the x values, plug in each of the solutions for y in the linear equation, x = 10 - 3y

When y = 1: x = 10 - 3(1) = 7
When y = 5: x = 10 - 3(5) = -5

Therefore, our two points of intersection are
(7, 1) and (-5, 5)

Answer 3

The Line 3y+X=10 intersetcs the circle x^2 + y^2 = 50 in two distinct points
x = 10 - 3y
(10-3y)^2 +y^2 = 50
100 -60y +9y^2 +y^2 = 50
10y^2 -60y +50 = 0
y^2 - 6y +5 = 0
(y-5)(y-1)=0 --> y= 5 and y=1
So x = 10 -3y = -5 or 7
So the points of interection are (-5,5) and (7,1)

Answer 4

It´s NOT a trig.question.You have a system of two equations

x^2+y^2-50=0 and x= 10-3y.By substitution

(10-3y)^2 +y^2-50=0 So 10y^2 -60y +50 =0

y^2-6y+5=0 y= (6+- 2)/2

y=4 y= 2 the corresponding x are x=-2 x= 4

The points are (-2,4) and (4,2)

Answer 5

There's a huge shortcut if you take a few seconds to guess integers.

Find two squares that add to 50:
A. 7^2 + 1^2 -- on the line when y=1 and x=7.
B. 5^2 + 5^2 -- on the line when y=5 and x=-5.

So the two answers stated by others are correct:
(7,1) and (-5,5).

Answer 6

Some help: solve for x in the first equation and substitute the result in the second equation. Solve for y using the quadratic formula, then use the two resulting values of y back in the first equation to find the corresponding x values.

This should help you.