After the coordinate origin and the point p 1,1 , and the center of the circle is on the line 2x 3y

Because the circle passes through the origin o(0,0) and p(1,1).

The slope of op k = (1-0) (1-0) = 1l and the slope of k = -1

Because l passes through the midpoint of the op (1 2, 1 2).

So the equation for l is y=1-x

Because the center of the circle is on l, and at the same time on the straight line 2x+3y+1=0.

So the center of the circle is the intersection of two straight lines (4,-3).

So let the equation for a circle be (x-4) +y+3) = r and substitute the origin o(0,0) for r = 16+9 = 25, so the equation for a circle is (x-4) +y+3) =25

The midpoint coordinates of the po (1 2, 1 2) and the slope are 1, so the perpendicular equation of the po is y-1 2=-1*(x-1 2), that is, y=-x+1

Coupled with 2x+3y+1=0, we get x=4 y=-3 so the center of the circle (4,-3) radius root number (4 2+3 2) = 5 so the equation of the circle (x-4) 2+(y+3) 2=25

Let (x-a) 2+(y-b) 2=r 2 substitute the two points.

a^2+b^2=r^2

a-1)^2+(b-1)^2=r^2

Derived. a+b=1

The center of the circle (a, b) is in a straight line.

Then 2a+3b+1=0

A system of equations for simultaneous solution of two formulas.

b=-3 a=4

Let the center of the circle c(x,y) satisfy oc=pc, that is, x 2+y 2=(x-1) 2+(y-1) 2, and 2x+3y+1=0, and the simultaneous solution yields x=4, y=-3Thus the radius r=5 and the circular equation is (x-4) 2+(y+3) 2=25

According to the geometric properties of the circle, let the origin point be 0, op is the chord of the circle, and the center of the circle should lie on the line perpendicular to the string, so the midpoint of op (1 2, 1 2), to be perpendicular to op, the slope is -1, so the equation of the circle y=-x+1 is in 2x+3y+1=0 and the center of the circle (4,-3),r=5 so the circle is (x-4) 2+(y+3) 2=25

The distance from the center of the circle (origin) to the bucket stool in a straight line = radius = |10|Thick Nagan number (4 2 + 3 2) = 2

So, the circular equation is: x 2 + y 2 = 4

Find the distance from the point to the straight line, and the distance you get is the radius of the round hand, and then you can make a circle.

The over-circular Zen equation x2+y2=r2

d=ax+by+c/a2+b2

Solution 1: Because the center of the circle is on the straight line y=x+2, let the coordinates of the center of the circle be (a a+2).

Then the equation for the circle is (x-a)2

y-a-2)2

r2 is <> because the points o(0 0) and p(1 3) are on a circle

The solution is <>

So the equation for the circle is (x+<>

y<>

Solution 2: The slope of the round chord OP is 3 and the coordinates of the midpoint are (<>

So the perpendicular bisector equation for the chord op is y<>

x<> i.e., x+3y-5=0

Because the center of the circle is on the straight line y=x+2 and the center of the circle is on the perpendicular bisecting line of the chord op, it is <>

The solution is <>

That is, the center of the circle sits as if the label is changed to c(<>

And because the radius of the circle r=|oc|=<>

So the equation for the circle is (x+<>

y<>

Comments: There are three quantities of a, b, and r in the standard equation of a circle, that is, three quantities of a, b, and r are required for a circle, and sometimes the undetermined coefficient method can be used. It is necessary to pay attention to the application of relevant knowledge in plane geometry in problem solving.

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

Center of the circle c = (m, 2m).

Crossing points a(-1,0) and b(1,-2).

ac|=|bc|

m+1)^2+(2m-0)^2 = m-1)^2+(2m+2)^25m^2+2m+1 = 5m^2+6m+54m=-4m=-1

c=(m,2m)=(1,-2)

r^2=|ac|2 = m+1) 2+(2m-0) 2 = 16 circles. (x+1)^2+(y+2)^2 =16

The equation of the straight line where the perpendicular bisector of the line segment passing through the origin (0,0) and p(1,1) is x y=1, then the intersection of this line and 2x 3y 1=0 is the center of the circle, and if the center of the circle is (4, 3), then the radius of the circle is r=5, and the circle equation is (x 4) y 3) =25

The midpoint of the points o(0,0) and p(1,1) is (1 2,1 2), and the slope of the straight line op is 1, then the perpendicular bisector passes through the point (1 2,1 2) and the slope is 1, that is: y= (x 1 2) 1 2, and simplification obtains: x y=1

Because the circle passes through the origin o(0,0) and p(1,1).

The slope of op k=(1-0) (1-0).

The slope of l is -1 k

Because l passes through the midpoint of the op (1 2, 1 2).

So the equation for l is.

y=1-x because the center of the circle is on l, and at the same time on the linear modulus 2x+3y+1=0.

So the center of the circle is the intersection of two straight lines.

So let the equation for a row or circle be (x-4) +y+3) r and replace the original file with o(0,0). The equation for the circle is .

x-4)²+y+3)²

The coordinate origin o and the point silver collapse a (

Midpoint of OA (1 2, 1 2).

The center of the circle C must be in a straight line perpendicular to OA through the midpoint of OA (1 2, 1 2).

And since the slope of the straight line OA is 1, the slope of the straight line perpendicular to OA should be -1 front circle. The equation for a straight line perpendicular to OA is.

x+y-1=0

The center of the circle is again in a straight line 2x

y0. Solve the system of equations: x+y-1=0xy

Get the center of the circle c(4,-3).

Find r=oc=5

So the circle c equation.

x-4）²y+3）²

Let the center of the circle (a, b) be the barricade, the radius be r, let the equation of the circle c be (state letter x-a) 2 + (y-b) 2 = r2 (where 2 is squared), and solve the system of equations: 2a + 3b + 1 = 0 (where 2 is 2); a2+b2=r2 (where 2 is squared); (1-a)2+(1-b)2=r2 (where the balance of 2 is squared).

The coordinates of the midpoint of the bending mu of PO (1 2, 1 2), and the slope is 1, so the perpendicular equation of PO is y-1 2=-1*(x-1 2), that is, y=-x+1

Coupled with 2x+3y+1=0, x=4 is obtained

y=-3, so the center of the circle (4, -3).

Radius root number (4 2 + 3 2) = 5

The buried forest is based on the equation of the circle (Jie Shan x-4) 2+(y+3) 2=25

First, the equation for the circle is (x-a) +y-b) =r (a,b) is the center of the circle.

Then substitute the coordinates of the origin and p points into a +b =r (1-a) +1-b) =r

i.e. a +b = (1-a) +1-b) =1-2a+a +1-2b+b =2-2a-2b+a +b

2-2a-2b=0

Substituting (a,b) into 2x+3y+1=0 to obtain the stove mold Zheng 2a+3b+1=0 bili 2-2a-2b=02a+3b+1=0 to obtain a=8, b=-3 and then substituting a +b =r to obtain r =73

The equation for the circle is (x-8) +y+3) code crack = 73