Find the equation for the garden, the center of the circle is on the y axis, the radius is 5, and it

The center of the circle is on the y-axis, so let the coordinates of the center of the circle be (0,a) and the radius is 5, so let the equation of the circle be x 2+(y-a) 2=5 2, substitute the ab coordinates into the equation respectively, and take the common solution to obtain a=6, so that the equation of the circle is x 2+(y-6) 2=25

The center of the circle is on the y-axis Let the coordinates of the center of the circle be (0,y) according to the equation of the distance from the center of the circle to any point on the circle.

0-(-3)] square + (y-2) square = (0-3) square + (y-10) square.

Solving the equation gives us y=6 and the radius is 5 according to the distance between the center of the circle and the point, so the equation for the circle is x-squared + (y-6) squared = 25

Solution: Let the circular equation be (x) 2+(y-y1) 2=25, pass the points a, b, and substitute y1=

So the center of the circle is (0,y1).

x^2+(y-8)^2=25

You let the equation be x 2+(y-b) 2=25 and substitute those two points to get b=8

Find the equation of the straight line ab so that x=0 is the center of the circle. It's simple, right?

Let the center of the circle be (x,0) (x-2) +9=25 x=-2 or 6

The circular equation is (x+2) +y =25 or (x-6) +y =25

Let the equation be (x-a) 2+y 2=r 2, then:

2-a)^2+(-3)^2=5^2

The solution yields a1=-2 and a2=6

The equation is (x+2) 2+y 2=25 or (x-6) 2+y 2=25

The center of the circle is the midpoint of the diameter, and the coordinates of the two points of the diameter (4,0)(0,-6) can be known by using the triangle similarity

Then the radius r = under the root number (4 squared + 6 squared) 2 = root 13

Then the circle: (x-2) 2+(y+3) 2=13

Set the round round fiber hidden pure heart is (carry block x, 0).

Then because the circle is tangent to the straight line x+2y=0.

So d=lx+2 0+0l (destroy 1 +2 )=5, so lxl=5 gets x= 5

Because the circle is on the left side of the y-axis, x=5

So the center of the circle is (5,0).

Then the equation for the circle (x-5) +y =5

Method 1:

Let the original equation be: (x-a).

y =r two points can be substituted into the solution of the system of equations.

Let the equation of the line where ab is located is: y=kx

m substitutes the two points a and b, and the solution is: k = 1

m=2 so the straight line ab is: y=x

The coordinates of the midpoint in 2ab are (0,2).

Let the line perpendicular to ab be: y=-x

N substitute (0,2) to get y=-x

2. This straight line passes through the center of the circle.

Let y=0x=2, so the coordinates of the center of the circle o are o(2,0).

r=ao=√（9

So the original equation is (x-2).

y²=10

The length of the line segment AB = root number (.)

The midpoint d( of the line segment AB is obtained from the midpoint coordinate formula

Because point d is on the x-axis and the center of the circle is on the x-axis, the center of the circle passing through a and b on the x-axis is d, and the radius is 5;

The equation for this circle: (x+2) 2+y 2=25

Since the center of the circle is on the x-axis, let the center of the circle be (x1,0), so the circle equation is (x x1) square y square r square, bring in the coordinates of a and b respectively, and obtain a system of equations, which can solve x1 4, r square 13, so the circle equation is (x

4) Square y, square 13

The length of the line segment AB = root number (.)

The midpoint d( of the line segment AB is obtained from the midpoint coordinate formula

Because point d is on the x-axis and the center of the circle is on the x-axis, the center of the circle passing through a and b on the x-axis is d, and the radius is 5;

The equation for this circle: (x+2) 2+y 2=25

Since the center of the circle is on the x-axis, let the center of the circle be (x1,0), so the circle equation is (x x1) square y square r square, bring in the coordinates of a and b respectively, and obtain a system of equations, which can solve x1 4, r square 13, so the circle equation is (x

4) Square y, square 13

Method 1:

Let the original equation be: (x-a).

y =r two points can be substituted into the solution of the system of equations.

Let the equation of the line where ab is located is: y=kx

m substitutes the two points a and b, and the solution is: k = 1

m=2 so the straight line ab is: y=x

The coordinates of the midpoint in 2ab are (0,2).

Let the line perpendicular to ab be: y=-x

N substitute (0,2) to get y=-x

2. This straight line passes through the center of the circle.

Let y=0x=2, so the coordinates of the center of the circle o are o(2,0).

r=ao=√（9

So the original equation is (x-2).

y²=10

Solution: Let the center of the circle be (0, rolling a).

5-0)²+2-a)²=3-0)²+0-a)²25+4-4a+a²=9+a²

4a=20a=5

Round heart wax early (0, 5).

Radius = (5-0) +2-5) Equation for a large Qi = 34 circle: x +(y-5) =34

Center of the circle (a,0).

r =25, so the rock burn (a-6) +0-3) =25a paragraph letter -12a + 36 + 9 = 25

a-2)(a-10)=0

So. x-2) +y = 25 and (x-10) gross limb +y = 25

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