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
the actual answer is 4
it is actually 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 $13-3-6 = \boxed{4}$.
Oh, I see you're asking for the shortest distance between two circles. Well, in the spirit of humor, I'll give you a funny answer:
Those circles must be really close friends if they want to know the shortest distance between them! If they're asking, maybe they want to plan a socially distanced party or something. Sorry, circles, but I can't help you measure your friendship distance.
To find the shortest distance between the two circles, we can follow these steps:
1. First, we need to find the center and radius of each circle. The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is its radius.
For the first circle with the equation x^2 - 10x + y^2 - 4y - 7 = 0, we complete the square for x and y separately:
(x^2 - 10x) + (y^2 - 4y) - 7 = 0
(x^2 - 10x + 25) + (y^2 - 4y + 4) - 7 = 25 + 4
(x - 5)^2 + (y - 2)^2 = 36
So, the center of the first circle is at (5, 2), and its radius is √36 = 6.
Similarly, for the second circle with the equation x^2 + 14x + y^2 + 6y + 49 = 0, we complete the square:
(x^2 + 14x) + (y^2 + 6y) + 49 = 0
(x^2 + 14x + 49) + (y^2 + 6y + 9) + 49 = 49 + 9
(x + 7)^2 + (y + 3)^2 = 19
Hence, the center of the second circle is at (-7, -3), and its radius is √19.
2. Now, we have the centers and radii of both circles. We can calculate the distance between their centers using the distance formula, given by:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Plugging in the coordinates, we get:
Distance = √((-7 - 5)^2 + (-3 - 2)^2)
Distance = √((-12)^2 + (-5)^2)
Distance = √(144 + 25)
Distance = √169
Distance = 13
Therefore, the shortest distance between the two circles is 13 units.
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