Find the equation for a circle with a meridian point A 3 2 B 2 1 and the center of the circle on the

I wonder, is your question a straight line x-2y-3=0? Otherwise, there will be a in the expression, but it will be much simpler, because it will become a straight line parallel to the x-axis. (I don't know if I can do it, I haven't tried it, and it seems that I can't do it).

So I'll do it with x-2y-3=0. Since it passes through points a and b, let the center of the circle be q(a,b), and the distance from q to a and b is equal, both of which are radius r, then the equation can be listed: (a-3) 2+(b-2) 2=(a-2) 2+(b-1) 2

After the parentheses on both sides are squared, you can get an equation: a+b=4( ) I don't know if it's right, because it's mentally arithmetic, hehe), because (a, b) the straight line x-2y-3=0, so you can list the equation: a-2b-3=0 ( ) can solve the inequality by , that is, get this point, and finally substitute it into the above equation, and solve r.

q：（11/3，1/3）r∧2=29/9

If you have (a, b) and r, you can find the equation for this circle.

It's x-2y-3=0, right?

i.e. x=2y+3

Let the center of the circle o(2a+3,a).

oa²=ob²=r²

i.e.: (2a+3-3) +a-2) =(2a+3-2) +a-1) =r

The solution yields a = 1 3 and r = 29 9

o(11/3,1/3)

x-11/3)²+y-1/3)²=29/9

The midpoint of ab c(1,1), the slope of ab.

Therefore, the slope of the perpendicular line in the peidou ab is -2

Therefore, the middle vertical line is changed to the core branch of the middle sensitive equation:

y-1=-2(x-1)

That is: 2x+y-3=0

Synoptic: x+2y=0

x=2y=-1

So the center of the circle is (2,-1).

r^2=3^2+1^2=10

The circular equation is: x-2) 2+(y+1) 2=10ps: use the perpendicular line of the string to pass through the center of the circle to find the center of the circle.

a(5,2),b(3,2), the oblique scale of the straight line ab is 2-2 5-3 =0, the vertical bisector of the straight line ab is perpendicular to the x-axis, and its equation is: x= 5+3 2 =4, and the linear line 2x-3y-3=0 is solved in conjunction with x=4, y= 5 3, that is, the center m coordinate of the circle is (4, 5 3), and the radius of the circle is r=|am|= 5-4 ) 2 +(2-5 ) 2 = 10 , then the equation for the round-style void is (x-4) 2 + (dust slip y- 5 3 ) 2 = 10

Solution: The center of the circle is on the straight line 3x-y-2 0.

Let the center of the circle m (m, 3m-2).

The circle crosses two points a(3,1) and b(-1,3).

AB midpoint n is (1,2), KAB -1 2

kmn＝（3m-4）／（m-1）

Kab KMN -1, ie.

1 2) (3m-4) (m-1) -1, solution m 2, then 3m-2 split plexus vertical 4

Center m(2,4).

The radius of the circle is ma Zheng Sheng 10

The circular equation is (x-2) 2 (y-4) 2 10

This question is not difficult to do with a closed grip, but it requires basic knowledge.

Let the center of the circle be (2a+3,a).

then the distance from the center of the circle to the two points is equal.

So (2a+3-2) +a+3) = (2a+3+2) +a+5).

2a+1）²+a+3）²=（2a+5）²+a+5）²4a+1+6a+9=20a+25+10a+25-20a=40

a=-2 so the center of the circle is: (-1, -2).

Radius = (2a+1) +a+3) = 10 so the equation is: (x+1) +y+2) =10

The center c of the circle is in the line l: x-2y-3=0

Let the abscissa of the center of the circle be a, then its ordinate is (a-3) 2 The equation of the circle is (x-a) +y- (a-3) 2] =r Substituting the points a(2,-3) and b(-2,-5) gives (2-a) +3-(a-3) 2] =r (-2-a) +5-(a-3) 2] =r The two-formula simultaneous solution is obtained.

a=-1r²=10

The circular equation is (x+1) +y+2) =10

Let the center of the circle be (2a+3,a) and the radius be r, then the circle equation is:

x-2a-3) 2+(y-a) 2=r 2 then there is: (2-2a-3) 2+(-3-a) 2=r 2(-2-2a-3) 2+(-5-a) 2=r 2 solution: a=-2

r^2=10

Circular equation: (x+1) 2+(y+2) 2=10

Solution: Let the center of the circle c(a,b), then 2a-b=3, b=2a-3 The equation for the circle: (x-a) 2+(y-2a+3) 2=cx=3,y=-2; x=5 and y=2 are substituted respectively to obtain:

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

Solution: a=-2, c=50, b=2a-3=-7 The equation for the circle: (x+2) 2+(y+7) 2=50

Let the center of the circle be (a,b) and the radius be r

Since the center of the circle is on 2x-y=3, b=2a-3 lets the equation for the circle be.

x-a）²+y-2a+3）²=r²

Since a and b are both on a circle, they both satisfy the equation of a circle

5-a）²+2-2a+3）²=r²①

3-a) +2-2a+3) =r - get:

5-a）²-3-a）²]5-2a）²-1-2a）²]=0

Using the squared difference formula, we get:

5-a-3+a）（5-a+3-a）+（5-2a-1+2a）（5-2a+1-2a）=0

The solution is a=2, and substituting b gives b=1

Substituting a=2 into the equation, we get r =10

Hence the equation for a circle is:

x-2）²+y-1）²=10

Let the coordinates of the center of the circle be (x,y).

With the characteristics of a circle, it can be seen that the distance from the center of the circle to any point on the circle is the radius, so (x-2) 2+(y+3) 2=(x+2) 2+(y+5) 2 gives 8x+4y+16=0

Coupled with x-2y-3=0.

The coordinates of the center of the circle are (-1, -2).

The equation for a circle is (x+1) 2+(y+2) 2=10

Solution: The slope of the straight line x-2y-3=0 is 1 2, then the slope of the straight line ab is -2 because a(2-3) and b (-2.).-5), the equation for the straight line ab is y=-2x-4

x=-1 y=-2 r²=10

The equation for the garden (x+1) +y+2) =10

Solution: Set: Center coordinates: (x0, y0).

Radius: r then there are: (x-x0) 2+(y-y0) 2=r 2 because: the center of the circle is in a straight line, then: y0=6*(x0)......Liang pants fierce.

And the circle passes through the point: (-2, -3) and the point: (2, -5) i.e. .

2-x0）^2+（-3-y0）^2=r^2……2-x0) 2+(-5-y0) socks 2=r 2......Lianli Oak Bridge, solution, x0=-1

y0=6*(-1)=-6

r= 10, i.e., (x+1) 2+(y+6) 2=10

If the circle passes AB, the center of the circle is on the perpendicular bisector of AB.

The midpoint of ab is c(1,3).

The slope of ab is (2-4) (3+1)=-1 2, so the slope of ab perpendicular bisector is -1 (-1 2)=2, so ab perpendicular bisector is y-3=2(x-1).

The center of the circle is again on the y-axis.

Then x=0, so y-3=-2

So the center of the circle d(0,1).

then r =|ad|=(-1-0) +1-4) =10, so x +(y-1) =10

If the circle passes AB, the center of the circle is on the perpendicular bisector of AB.

The midpoint of ab is c(1,3).

The slope of ab is (2-4) (3

So the slope of the ab perpendicular bisector is -1 (-1 2)=2, so the ab perpendicular bisector is y-3=2(x-1).

The center of the circle is again on the y-axis.

Then x=0, so y-3=-2

So the center of the circle d(0,1).

then r =|ad|=(-1-0)

So x(y-1)=10

Adopt the next ha, thank you.

The center of the circle is on the perpendicular line, ab:

y=-1/2x+7/2

The midpoint is (1,3).

So the mid-vertical line is y-3=2(x-1).

Find the intersection of the mid-perpendicular line and the y-axis.

i.e. x=0, then y=1

So the center of the circle is zhidao(0,1).

The radius is the distance from the center of the circle to a or b = root number 10

So the equation is x +(y-1) = 10

Let the center of the circle be (0,y), and get (0+1)zd178; +(y-4) =(0-3) +y-2) gives y=1

radius =(0-3) +y-2) =10

The equation is x +(y-1) = 10