It is known that the center of the circle A x 1 2 y 2 16 is A, and the point B is on the circle A

Analysis: According to the conditions, the intangent phase of the fixed circle and the moving circle is determined, and then the equation of the curve c is written according to the definition of the ellipse.

Classification discussion, when y0=0, the test line l: 3x0x+4y0y-12=0 and the curve c have and only one intersection point, when y0≠0, the line and the elliptic equation are used as the point on the curve c, and there is only one solution, and the line and the ellipse have only one intersection point.

Answer: Solution:(i) The center of the circle a is a(1,0) and the radius r1=4,

Let the center of the circle m (x,y) and the radius be r2, according to the meaning of the title, r2=|mb|．

by ab|=2, we know that the point b is in the circle a, so that the circle m is tangent in the circle a,

Hence ma|=r1-r2, i.e. ma|+|mb|=4，

So, the trajectory of the point m is an ellipse with a,b as the focus

The straight line l and the curve c have only one intersection point (2,0).

When x0=-2, y0=0, the equation for the straight line l is x0=-2,

The straight line l and the curve c have only one intersection point (2,0).

Therefore, the straight line l and the curve c have and have an intersection point p(x0,y0),

In summary, the line l and the curve c have only one intersection point, and the intersection point is p(x0,y0).

1) Using the substitution method, let d(x,y), then b(2x-3,2y), and bring the coordinates of b into the equation of circle a(2x-3+1) 2+(2y) 2=16, that is, the d trajectory equation.

2) Because f is in the perpendicular bisector of be, so bf=ef, so af+ef=af+bf=ab=4, so f trajectory is elliptical, 2a=4, a=2, c=1, so f trajectory equation is x 2 4 +y 2 3 = 1.

Summary. Because it is tangent to the y-axis, the radius r is, and the distance from the center of the circle to the y-axis is 1, so the circle equation is (x 1) +y 0) 1 x y 2x 0x y ax b 0, so b 0, a 2

5.If the center of the circle +x 2+y 2-ax+b=0+ is at (1,0), and.

5.If the center of the circle +x 2+y 2-ax+b=0+ is at (1,0) and tangent to the y-axis, then the value of b.

Because it is tangent to the y-axis, the radius r is, and the distance from the center of the circle to the y-axis is 1, so the circular equation is the stool code key (x 1) +y 0) 1 x y 2x 0x y ax b 0, so b 0, a 2

There are two types of equations for extended circles, which are divided into standard equations and general equations. The standard equation for a circle is in the form of: (x-a) 2+(y-b) 2=r 2.

The general equation form of a circle is: x 2 + y 2 + dx + ey + f = 0. Compared with the standard equation, in fact, d = -2a, e = -2b, f = a 2 + b 2-r 2.

Summary. Hello, please add the question**, it is best to take a picture of the question, so as to better answer for you.

12.The known points (a,b) are circles x 2+y 2-4x-8y+16

Hello, please add the question**, it is best to take a picture of the question, so as to better answer for you.

x 2 + y 2 = a 2 The radius of the circle is a, and the lead distance from the center of the circle to the straight line xx+yy=a 2 is the mountain noise:

a^2|/√x^2+y^2)

Circle c1x2

Ten Y216 is centered on C1

Ridge 0,0), radius 4, circle c2

x-a) 2y21 center c2

a,0), with a radius of 1,c1

c2=|a|, circle c1x2 ten y2

16 with a round let c2

x-a) 2y21 tangent years of the bureau, |a|=4+1=5 or |a|=4-1=3 i.e. a= 5 or 3

So the answer is 5 or 3

The center of the circle is b(-1,0) and the radius is 4

Because the intersection point P of the AC perpendicular bisector and the line segment CB, so pc=ap, so bp+ap=bp+pc=bc=4, so the trajectory is elliptical.

So 2a = 4 a = 2 c = 1

Then b = a -c = 3

So the trajectory of the intersection p x 4 + y 3 = 1

Let m(a,2-a).

The equation for the circle is set to (x-a) +y-2+a) =r and substituting a(1,-1),b(-1,1), into it.

Get (1-a) +3+a) = r

1-a）²+1+a)²=r²

The solution yields a=1 and r=2

From the nature of the circle, it can be known that the center of the circle is on the perpendicular bisector of the ** segment AB, and the center of the circle is connected on the straight line L, and the coordinates of the center of the circle can be known to be (3,-2), and then the radius of the circle can be known to be 5, and then the distance from the center of the circle to the straight line X-Y+5=0 can be obtained as the root number 2 5 times of 5 times, and it can be known that the intersection point of the vertical line segment and the circle from the center of the circle to the straight line X-Y+5=0 is the P point that is sought, and the minimum value of PQ is 5 times the root number 2 minus 5, and the maximum value is 5 times the root number 2 plus 5

3 questions, from the above can know the coordinates of the center of the circle and the radius of the circle, through the center of the circle to the straight line kx-y+5 = 0 to do the perpendicular line segment from the circle to the straight line The distance formula can be expressed by k, and then connect the center of the circle and the intersection of the line and the circle to form a right triangle, the column equation is solved to obtain k = 20 21, find.