Question

What is the shortest distance between the circles defined by x^2-10x +y^2-4y-7=0 and x^2+14x +y^2+6y+49=0?

Thank you

Answer 1

find the centers and radii

x^2 - 10 x = -y^2 + 4 y +7

x^2 - 10 x + 25 = -y^2 + 4 y + 32

(x-5)^2 +y^2 -4 y + 4 = 36
(x-5)^2 + (y-4)^2 = 6^2
center at (5,4) r = 6

second one
x^2 + 14 x +49 + y^2 + 6 y = 0
(x+7)^2 + y^2 + 6 y + 9 = 9
(x+7)^2 + (y+3)^2 = 3^2
center at (-7,-3), r = 3
x between centers = 7+5 = 12
y between centers = 3+4 = 7
distance between centers
= sqrt (144+49)
= sqrt 193
subtract sum of radii
sqrt (193 ) - 3 -6
sqrt (193) - 9
4.9
CHECK MY ARITHMETIC. I have a hunch it should be a simpler right triangle in x and y

Answer 2

the actual answer is 4

Answer 3

it is actually 4

Answer 4

We complete the square for the first equation by observing that the first equation is equivalent to\[
(x^2-10x +25) +(y^2-4y +4)=36,
\]which is also equivalent to\[
(x-5)^2 +(y-2)^2 =6^2.
\]Similarly, the equation for the second circle is\[
(x+7)^2 +(y+3)^2 =3^2.
\]Hence, the centers of the circles are $(5,2)$ and $(-7,-3)$, and the radii of the circles are equal to 6 and 3, respectively. The distance between the points $(5,2)$ and $(-7,-3)$ by the distance formula is $\sqrt{(5-(-7))^2+(2-(-3))^2}=\sqrt{12^2+5^2}=\sqrt{169}=13$. Therefore, to find the shortest distance between the two circles, we must subtract from $13$ the sum of the radii of the two circles. Thus, the shortest distance between the circles is $13-3-6 = \boxed{4}$.