Given that the moving circle is circumtangent to the circle (x + 2) 2 + y2 = 4 and tangent to the straight line x = 2, the trajectory equation of the center P of the moving circle is obtained

Given that the moving circle is circumtangent to the circle (x + 2) 2 + y2 = 4 and tangent to the straight line x = 2, the trajectory equation of the center P of the moving circle is obtained

Let the center C1 (- 2, 0) of the circle (x + 2) 2 + y2 = 4, the center P (x, y) of the moving circle P, R, be x = 4, x = 2, PQ ⊥ the straight line x = 4, Q be the perpendicular foot. Because the circle P is tangent to x = 2, the distance from the circle p to the straight line x = 4 PQ = R + 2, and PC1 = R + 2, so the distance from P (x, y) to C1 (- 2, 0) and the straight line x = 4

The equation of the circle whose center is C (3, - 5) and tangent to the line x-7y + 2 = 0 is______ ．

∵ the distance from the center of a circle to the tangent d = R, that is, r = D = | 3 + 35 + 2 | 12 + 72 = 42, the center of a circle C (3, - 5), the equation of circle C is (x-3) 2 + (y + 5) 2 = 32. So the answer is: (x-3) 2 + (y + 5) 2 = 32

Given the circular equation x & # 178; + Y & # 178; = 1, the linear equation is y = X-B, when B is at what value, the circle and the line intersect? Tangent? Separation? Need to solve the problem process

Taking y = X-B into the equation of circle, we can get: 2x ^ 2-2bx + B ^ 2-1 = 0, so the problem of intersection of line and circle is transformed into the number of solutions of the equation, that is, when the equation has one solution, the circle is tangent to the line, if there are two solutions, the circle intersects with the line, if there is no solution, the circle is separated from the line. Because △ = - 4B ^ 2 + 8 of the equation, when x = radical 2, it is tangent, when x is greater than radical 2, it is separated, and when x is less than radical 2, it intersects