Find the points of intersection of x^2 + y^2 = 4 and (x-4)^2 + y^2 =12 ?

show that these circles intersect at right angles; that is, that their tangents at either point of intersection are perpendicular.





i found their equation for tangent slope:





y^2 + x^2 = 4 came out to be -x/y and (x-4)^2 + y^2 = 12 came out to be (-x+4)/y soo... what do I do to find the intersection points and how do I find the slopes of them to find out if they are perpendicular (reciprocal and opposite sign)?

Find the points of intersection of x^2 + y^2 = 4 and (x-4)^2 + y^2 =12 ?
x^2 + y^2 = 4








(x-4)^2 + y^2 =12


x^2 - 8x + 16 + y^2 = 12


(x^2 + y^2) - 8x + 16 = 12


4 - 8x + 16 = 12


20 - 8x = 12


-8x = -8


x = 1





When x = 1, from the first equation, 1 + y^2 = 4 or y = +/- sqrt(3)





Validating these two points in the second equation


(x-4)^2 + y^2 =


(1-4)^2 + (+/- sqrt(3))^2 = 9 + 3 = 12





So the two interesection points are (1, sqrt(3)) and (1, -sqrt(3))





Now for two lines to be perpendicular, their slopes need to be negative reciprocals...





I'll leave that to you to solve.
Reply:Ok, here we go!





Formulae 1. x^2+y^2=4


Formulae 2. (x-4)^2+y^2=12





Expand formulae 2.


[x^2-2(4x)+4^2]+y^2=12


[x^2-8x+8]+y^2=12


x^2-8x+y^2=12





Therefore, formulae 2 is x^2-8x+y^2=12





Formulae 1. x^2+y^2=4


Formulae 2. x^2-8x+y^2=12


The point of intersection MUST fit into this equation right?


So you can either find x first or y.





Formulae 1- Formulae 2


-8x=-8


Therefore, x equals 1.


Now substitute x into any Formulae you want. 1 is the best.


8^2+y^2=4


64+y^2=4


y^2=-60


y= Square root of -60.





Hope this helps! Good luck!