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Find the trajectory equation of the center of a circle which passes through point a (2,0) and is inscribed with circle x2 + 4x + y2-32 = 0
(x + 2) 2 + y2 = 36, (x + 2) 2 + y2 = 36, (x + 2) 2 + y2 = 36, (x + 2) 2 + y2 = 36, (x + 2) 2 + y2 = 36, (x + 2) 2 + y2 = 2 + y2 = 36, (x + 2 + 2) center coordinates of the circle x2 + 4x + y2-2-32 = 0 are (- 2, 0), radius is 6, and radius is 6 (x, y) the center coordinates of the moving circle is (x (x, y) the point a (2,0) passing through point a (2,0) and is inscribed with the circle x2 + 4x + 4x + 4x + Y2 + y2-2 + x2 = 6 {(x + 2) 2) 2 + 6 − + 4x + 4x + 4 + 4 + Y2 + 4 + Y2: the two sides of the two sides of the two sides of the square: x2 − − − \it's not easy The formula is: 5x2 + 9y2 = 45, that is, X29 + Y25 = 1
Given m (0, √ 3), the moving circle I passes through the point m and is inscribed with the circle n: X & # 178; + (y + √ 3) &# 178; = 16. (1) the equation of the locus C of the center of the circle I is solved
M（0,√3）,N（0,-√3）,
Let the radius of the moving circle be r. since circle I is inscribed with circle n, so | in | = 4-R = 4 - | im |,
So | in | + | im | = 4,
By definition, the trajectory of I is an ellipse with m and N as the focus,
2A = 4, a = 2, C = √ 3, so B ^ 2 = a ^ 2-C ^ 2 = 1,
The focus of the ellipse is on the Y axis, so the trajectory equation of I is y ^ 2 / 4 + x ^ 2 = 1
The moving circle is inscribed with the known circle O2: (X-2) &# 178; + Y & # 178; = 81, and is circumscribed with the known circle O1: (x + 2) &# 178; + Y & # 178; = 1,
Find the trajectory equation of moving circle C
Answer: (X-2) &# 178; + Y & # 178; = 81, center of circle is (2,0), radius r = 9 (x + 2) &# 178; + Y & # 178; = 1, center of circle is (- 2,0), radius r = 1, let radius of moving circle be m, center of moving circle be (x, y), then center distance of circumscribed circle = 1 + m > 1, center distance of inscribed circle = 9-m > 0, so: √ [(x + 2) &# 178; + Y & # 178;] = 1 + m √ [(x
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