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!
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