If the circles xsquare+ysquare-2x+4y-4=0 & xsquare+ysquare-8x-12y+c=0 touch each other. Find c.?

Answer 1

centre of the first circle is (-g,-f)=(1,-2)
center of the second circle=(4,6)
(distance between the centres) i.e.=[(4-1)^2+(6+2)]^2=(73)^1/2
(radius of the first circle)=(g^2+f^2-c)^1/2=3
(radiusof the second circle)=(52-c)^1/2
3+(52-c)^1/2=73^1/2 , squaring
9+52-c+6(52-c)^1/2=73
-12-c=-6(52-c)^1/2
144+c^2+24c=36(52-c) =>
c^2-12c-1728=0 =>(c-48)(c+36)=0 giving c=48

Answer 2

I assume that you mean that the circles are to be tangent to each other.

Find the center and radius of the 1st circle.

x^2 -2x +1 + y^2 +4y + 4 = 4 + 1 + 4 = 9 (done by completing the square, etc.)
(x-1)^2 + (y + 2)^2 = 9

C(1,-2) r = 3

Next, do the same thing for 2nd circle.

x^2 - 8x + 16 + y^2 +12 + 36 = -c + 16 + 36 = 52 - c
(x - 4)^2 + (y + 6)^2 = 52 - c

C(4,-6) r = sqrt(52 - c)

Now find the distance between the centers.

d = sqrt[(1-4)^2 + (-2 + 6)^2]
d = sqrt(9 + 16) = sqr(25) = 5

Since the circle are tangent and the radius of the 1st circle is 3, then the radius of the 2nd circle must be 2.

So, sqrt(52 - c) = 2
52 - c = 4
c = 48

Bingo

Answer 3

the value of c = -4

Answer 4

middle

Answer 5

underneath the doormat,or was that key?