The circle A is tangent to the y axis, and the center of the circle is on the straight line x 2y 0

Since the center of the circle is on the line x=2y, and above the x-axis, let the coordinates of the center of the circle be (2u,u), u>0

The circle is tangent to the y-axis, r=2u

The circular equation is. x-2u)^2+(y-u)^2=4u^2

Let y=0x-2u) 2=3u2

x-2u=±√(3)u

x1=(2-√3)u

x2=(2+√3)u

x2-x1=2√3)u=2√3

u=1 The circular equation is.

x-2)^2+(y-1)^2=4

2) Since the tangent of the circle passing point (4,4) is parallel to the x-axis or y-axis, the tangent equation is.

x = 4 and y = 4

The tangent length is 2.

3) Let the linear equation be.

y-2=k(x-1)

y=kx-k+2

Generational circle equations.

x-2)^2+(kx-k+1)^2=4

1+k^2)x^2+(2k-2k^2-4)x+(k-1)^2=0

x1+x2=(2k^2-2k+4)/(1+k^2)

x=(k^2-k+2)/(1+k^2)

1+k^2)x=k^2-k+2

Replace k=(y-2) (x-1) with the above formula.

1+(y-2)^2/(x-1)^2]x=(y-2)^2/(x-1)^2-(y-2)/(x-1)+2

x(x-1)^2+x(y-2)^2=(y-2)^2-(y-2)(x-1)+2(x-1)^2

x(x^2-2x+1)+x(y^2-4y+4)=y^2-4y+4-xy+2x+y-2+2x^2-4x+2

x^3-4x^2+9x+xy^2-3xy-y^2+3y-4=0

The line connecting the center of the circle and the tangent point is perpendicular to the tangent line, so when it is tangent to the y-axis, the line between the center of the circle and the tangent point is perpendicular to the y-axis.

Perpendicular to the y-axis, the straight line passing through (0,1) is y=1

So the intersection of y=1 and x-2y=0 is the center of the circle.

The distance between the center of the circle (2,1) and the (0,1) god is the radius of You.

So radius = 2

x-2)^2+(y-1)^2=4

Let the equation for this circle be: (x-a) 2+(y-b) 2=r 2, then the center of the circle is (a,b) and the radius is r, and the center of the circle is obtained on the line x-3y=0: a-3b=0......From this circle passes through the point a(6,1).

6-a)^2+(1-b)^2=r^2……And the circle is tangent to the y-axis, then the radius of the circle is equal to the abscissa of the center of the circle, i.e.: r=|a|……

It is tangent to the blind coordinate axis, so the distance from the center of the circle to the axis of the two sitting finches is equal, so x=y or x=-y

The center of the circle is on the straight line x+2y+3=0.

If x=y, then x=y=-1;If x=-y, then x=3, y=-3, so the center of the circle is (-1, -1) or (3, -3).

Because the radius is the distance from the center of the circle to the tangent, that is, the distance to the coordinate axis.

So the center of the circle is (-1, -1), then r=1;If the center of the circle is (3,-3), then r=3, so the standard equations for the circle are (x+1) 2 +(y+1) 2 =1 and (x-3) 2 +(y+3) 2 =9

So the answers are: (x+1) 2 +(y+1) 2 =1 and (x-3) 2 + (no y+3) 2 =9

Solution: Since the center of the circle is on x-3y=0, let the coordinates of the center of the circle be (3m,m) and m 0, and the radius of 3m is obtained according to the tangent of the circle and the y-axis

Then the equation of the circle is (x-3m)2+(y-m)2=9m2, and substituting a(6,1) into the equation of the circle obtains: return to the tomb (leak Ziqi 6-3m) 2+(1-m)2=9m2, Qi leak.

Simplified: m2-38m+37=0, then m=1 or 37, so the equation for the circle is (x-3)2+(y-1)2=9 or (x-111)2+(y-37)2=12321

If the center of the limb circle is (3a,a), then the radius is 3|a|，3a-6)^2+(a-1)^2=9a^2

a^2-38a+37=0

a = 1 or a = 37

When a=1, the center of the circle is (3,1), the radius is 3, and the equation is.

x-3)^2+(y-1)^2=9

When a=37, the punch rule.

The center of the circle is (111,37), and the radius of the scattered shed is 111, and the equation is. x-111)^2+(y-37)^2=12321

Since the center of the circle is on the straight line x-3y=0, let the center of the circle (3a,a), and the square of the circle is (x-3a) 2+(y-a) 2=r 2

tangent to the y-axis, so the radius r=3 |a|

Circle or pass through point A (6,1).

Therefore, (6-3a) 2+(1-a) 2=r 2 two equations and two unknowns, find a,r

Are there any extra points? Blocking branches.

Let the coordinates of the center of the circle be o(3b,b), because the circle is tangent to y, so the radius of the circle is |3b|, and because the circle passes through the point a(6,1), so, (3b-6) 2+(b-1) 2=|3b|^2

The solution yields b=1 or b=39

The equation for the circle is (x-3) 2+(y-1) 2=9x-117) 2+(y-39) 2=117 2

Let the circle equation (x-a) 2+(y-b) 2=r 2 The circle is tangent to the y-axis, a|=r

The center of the circle is in the line x-3y=0, a=3b

And this circle crosses a(6,1),(6-a) 2+(1-b) 2=r 2(6-3b) 2+(1-b) 2=9b 2b 2-38b+37=0

According to the title, the distance from the center of the circle to the y-axis is the same as the distance to point a.

The center of the circle is on the straight line x-3y=0, so let the center of the circle be: (3a,a)

The distance from the center of the circle to the y-axis of the rock wax is |3a|

The square of the distance from the center of the circle to point A is (3a-6) 2+(a-1) 29a 2=(3a-6) 2+(a-1) 2, giving a=1 or 37

When a=1, the radius is 3, the center of the circle is (3,1), and the equation is (x-3) 2+(y-1) 2=9

When a=37, the radius of the early jujube hand is 111, the center of the circle is (111,37), and the equation is (x-111) 2+(y-37) 2=111 2