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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

 Nov 29, 2023
 #1
avatar+671 
0

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.

 Nov 29, 2023
 #2
avatar+36923 
+3

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:

 Nov 29, 2023
edited by ElectricPavlov  Nov 29, 2023

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