two circles whose equations are (x-3)^2+(y-5)^2=25 and (x-7)^2+(y-5)^2=9 intersect in two points. What is the equation of the line passing through these two points?
Can anyone please give me some ideas how to do it without graphing it out? THANKS A LOT!
The centers are at (3,5) and at (7,5) along the horizontal line y = 5 and 7-3 = 4 apart
left radius = 5
right radius = 3
so
the points lie on vertical line x = something an equal distance above and below y = 5
so
triangle
base = 4 along y = 5 line
left leg = 5
right leg = 3
let's use law of cosines to find the angle on the left, between the y = 5 line and the long side length 5
3^2 = 5^2 + 4^2 - 2*5*4 cos A
9 = 25 + 16 - 40 cos A
32 = 40 cos A
A = 36.87 degrees
now the distance from x = 3 to our desired vertical line is
5 cos A which is 5 (32/40) = 4
so our line is
x = 3 + 4 = 7
which we could have said earlier seeing that our triangle is
3, 4, 5 and therefore a right triangle
What do you mean?
What do youm mean?
To find the equation of the line passing through the intersection points of the two circles, you need to determine the coordinates of these points first. Here's how you can do it:
1. Solve the system of equations formed by the two circle equations:
(x - 3)^2 + (y - 5)^2 = 25 ... Equation 1
(x - 7)^2 + (y - 5)^2 = 9 ... Equation 2
Expand the equations and simplify:
x^2 - 6x + 9 + y^2 - 10y + 25 = 25 ... Equation 1
x^2 - 14x + 49 + y^2 - 10y + 25 = 9 ... Equation 2
Combine like terms:
x^2 - 6x + y^2 - 10y + 34 = 0 ... Equation 1
x^2 - 14x + y^2 - 10y + 15 = 0 ... Equation 2
Now you have a system of equations with two variables (x and y). Subtract the second equation from the first, which eliminates the x^2 and y^2 terms:
(x^2 - 6x) - (x^2 - 14x) + (y^2 - 10y) - (y^2 - 10y) + (34 - 15) = 0
Simplify and combine like terms:
8x - 14x = -1
-6x = -1
x = 1/6
Substitute the value of x into either equation (preferably Equation 1) to solve for y:
(1/6)^2 - 6(1/6) + y^2 - 10y + 34 = 0
1/36 - 6/6 + y^2 - 10y + 34 = 0
1/36 - 36/6 + y^2 - 10y + 34 = 0
-6 + y^2 - 10y + 34 = 0
y^2 - 10y + 28 = 0
Solve this quadratic equation to find the values of y. The solutions will be the y-coordinates of the two intersection points.
2. Once you have the coordinates of the two intersection points, you can use the two-point form of the equation of a line to find the equation. The two-point form is:
(y - y1) = ((y2 - y1) / (x2 - x1))(x - x1)
Substitute the coordinates of one intersection point as (x1, y1) and the other point as (x2, y2). Then, substitute these values into the equation and simplify to obtain the equation of the line passing through the intersection points.
Remember to simplify and check your calculations along the way to ensure accuracy.