Question

There are two circles with the equations x^2+(y-4)^2=10 and (x-8)^2+y^2=50. Would the two circles intersect? If so, at what points?

Answer 1

First expand both equations:
\[x ^{2}+y ^{2}-8y+16-10=0\] and \[y ^{2}+ x ^{2}-16x+64-50=0\]. Now subtract both equations to have:\[-8y+6-(-16x+14)=0\] which is\[-8y =-16x+8\] which simplifies to\[y=2x-1\]. Now substitute this same y in one of the two equations, \[x ^{2}+(2x-1-4)^{2}=10\] which gives you\[x^{2}+(2x-5)^{2}=10\]. Expanding we have \[x ^{2}+4x ^{2}-20x+25=10\] which give you\[5x^{2}-20x+15=5(x ^{2}-4x+3)=5(x-3)(x-1)\]. Therefore both circles intersect at (3,5) and (1,1). Peace, Love and Happinnes from Puerto Rico

Answer 2

Solve each equation for y^2:
\[y^{2} = 8y-x^{2}-6\]
\[y^{2} = -x^{2}+16x-14\]
Set them equal to each other and solve for y:
\[8y -x^{2}-6 = -x^{2}+16x-14\]
\[8y = 16x -8\]
\[y =2x-1\]
This is the line that will go through any intersection points
Substitute this in for y into one of original equations
\[(x-8)^{2} +(2x-1)^{2} = 50\]
\[(x^{2}-16x+64)+(4x^{2}-4x+1) = 50\]
\[5x^{2}-20x+15 = 0\]
\[x^{2}-4x+3 = 0\]
\[(x-3)(x-1) = 0\]
\[x = 1, 3\]
Substitute into y=2x-1 to find corresponding y_values
\[y =2(1)-1 = 1\]
\[y=2(3) -1 = 5\]
Two intersecting points are (1,1) and (3,5)

Answer 3

Math never lies...

Answer 4

haha we did it pretty much the exact same way
very true...

Answer 5

Thank you~ I had this question on my math test but I got it wrong... so I was just curious to see how it could be done. I wish I had both of your brains! LOL