类型 Algebra 问题 What is the shortest distance between the circles defined by $x^{2}-24 x+y^{2}-32 y+384=0$ and $x^{2}+24 x+y^{2}+32 y+384=0 ?$ 标准答案

Step 1 ：Complete the square for the first equation by adding \((-24/2)^2\) and \((-32/2)^2\) to both sides, which gives \[(x^2-24x +144) +(y^2-32y +256)-16=0,\] which is also equivalent to \[(x-12)^2 +(y-16)^2 =4^2.\]

Step 2 ：Similarly, the equation for the second circle is \[(x+12)^2 +(y+16)^2 =4^2.\]

Step 3 ：Hence, the centers of the circles are \((12,16)\) and \((-12,-16)\) respectively. Furthermore, the radii of the circles are equal to \(4\).

Step 4 ：Now the distance between the points \((12,16)\) and \((-12,-16)\) by the distance formula or similarity of \(3-4-5\) triangles is \(40\).

Step 5 ：Therefore, to find the shortest distance between the two circles, we must subtract from \(40\) the distances from the centers to the circles. Thus, the shortest distance between the circles is \(40-4-4\).

Step 6 ：\(\boxed{32}\)