+180 The circles x^2+y^2=4 and (x-5)^2+(y-8)^2=57 intersect in two points A and B Find the slope of line AB
+743 Sure, here's how to find the slope of line AB:
Rewrite each circle equation in standard form:
x^2 + y^2 = 4 => (x^2 - 4) + y^2 = 0
(x - 5)^2 + (y - 8)^2 = 57 => (x^2 - 10x + 25) + (y^2 - 16y + 64) = 0
Add the two equations: 2x^2 - 10x + y^2 - 16y = 39
Group like terms: 2(x^2 - 5x) + y^2 - 16y = 39
Factor out the common factors: 2(x - 5)(x - 5) + y(y - 16) = 39
Solve for y: y = (39 - 2(x - 5)(x - 5)) / (y - 16)
To find the slope of line AB, we need to find the coordinates of points A and B. We can do this by substituting different values of x into the equation for y and solving. For example, if we substitute x = 5, we get y = 11. Therefore, point A is (5, 11).
Similarly, if we substitute x = 8, we get y = 0. Therefore, point B is (8, 0).
Calculate the slope of line AB: (11 - 0) / (5 - 8) = 2
Therefore, the slope of line AB is 2.
+37146 We are looking for the two points where the equations intersect....where they are equal ...so equate the two circle equations
x^2 + y^2 -4 = ( x-5)^2 + (y-8)^2 - 57 expand ...and then simplify to
0 = -10x -16y+36 which rearranges to
y = - 10/16 +36/16
y = - 5/8 x +9/4 this is the equation of the line AB......the slope is - 5/8
Here is the DESMOS graph showing the two circles and the (green) line:
