It is known that the circle passes through two points, P 4, 2 and Q 1, 3 , and the length of the li

Solution: Let the equation for the circle be: (x-a) 2+(y-b) 2=r 2 from the known conditions.

x=00-a)^2+(y-b)^2=r^2y=b±√（r^2-a^2）

y1-y2|=2√（r^2-a^2）=4√3...1) The garden passes through the two points of p(4,-2) and q(-1,3), and substitutes the coordinates of these two points into the park equation to get the next equation:

4-a)^2+(-2-b)^2=r^2...2)-1-a)^2+(3-b)^2=r^2...3) Solve the above equations (1), (2), (3) to obtain.

a=1,5,b=0,4,r 2=13,37 substitution of the kindergarten equation, obtain.

x-1)^2+y^2=13

x-5)^2+(y-4)^2=37

Tested to meet known conditions.

Answer: There are two equations for a circle: (x-1) 2+y 2=13 or (x-5) 2+(y-4) 2=37

Solution: Let the equation of the circle be x 2 + dx + y 2 + ey+ f = 0, and substitute the two points a(4,-2) and b(-1,3) into the equation to obtain.

16+4d+4-2e+f=0

1-d+9+3e+f=0

The length of the line segment on the y-axis is 4 and 3

48 + 4 roots 3e + f = 0

Solution: d, e, f.

Let the connoyance equation of the circle be (x-a) 2+(y-b) 2=r 2.

Substituting the two points p(4,2) and q(-1,3) into them gives two square laxiao cheng.

Then let x=0,(y-b) 2=r 2-a 2 so that the two values of y are solved, and the large one is subtracted from the small one is equal to 4, that is, r 2-a 2=4 can be solved by 3 equations.

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In fact, the equation for the circle is as: (x-a) +y-b) = r The above equation is: x=0, and we get: y= r 2-a 2) + b, which is the intersection of the y-axis of the circle when x=0.

r 2-a 2)+b)-(r 2-a 2)+b) circle on the y-axis of the line segment is 4 3 long

2√(r^2-a^2)=4√3 ①

Substituting the coordinate values of the two points p and q into the hypothetical circular equation yields:

4-a)²+2-b)²=r² ②

1-a)²+3-b)²=r² ③

The above three agendas can be solved together, and the values of a, b, r, can be obtained, and the obtained values are substituted into the set circular equation, that is, the branch of the circular equation is found.

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I'll tell you a way to do it yourself. Let the coordinates of the center of the circle be o(c,d) because op oq thus gives a unary linear equation.

Then let the intersection points of the circle and the y-axis be a and b, then the point A is (0, m), then the point b (0, m+4 * root number 3), and divide the trigonometric late lift form OAB into two equal and right-angled triangles, and the OA square ob square Qingbi can be found by solving the above equations.

In the process, the coordinates of the center of the circle should be set to (a, b).

According to "and the line intercepted on the y-axis contains four long mountain sections and number three", the mountain difference is x=0, which indicates the intersection coordinates of the circle and the y-axis.

y= r 2-a 2)", r is the radius, a is the abscissa of the center of the circle, and the distance from the center of the circle to the y-axis, and the y=b line is the section of the circle and the y-axis perpendicularly.

Just draw a sketch and see.

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Solution: Let the equation for the circle be (x-a) 2+(y-b) 2=r 2 (4-a) 2+(2+b) 2=r 2

1+a)^2+(3-b)^2=r^2

Let x=0,y= (r 2-a 2)+b

2√(r^2-a^2)=4√3

The joint solution is a=b+1

4-a) 2+(a+1) 2=a 2+12 a=1, (rounded) a=5

b=a-1=4,y^2=25+12=37

x-5)^2+(y-4)^2=37

Let the equation for the circle be (x-a) 2+(y-b) 2=r 2

Substituting p(4,-2) and q(-1,3) into the two equations. Substitute x 0 for |y1-y2|=4 3, let's solve the system of equations.

Substituting p q to get the equation:

a+1）^2+(3-b)^2=r^2

Let x=0y2-2yb+4b-12=0

According to Veda y1 + y2 = 2b

y1*y2=4b-12

Then y1-y2=y1+y2) 2-4y1y2-4 root number 34b 2-4*(4b-12)=48

b=0 a=1 r^2=13

The circular equation is. x-1)^2+y^2=13

Let the equation for the circle be (x-a) +y-b) =r (4-a) +2+b) =r , 1+a) +3-b) =r so that x=0 and y= (r -a )+b

2√(r²-a²)=4√3

The joint solution is a=b+1

4-a)²+a+1)²=a²+12

a=1, (rounded) a=5

b=a-1=4,r²=25+12=37

x-5)²+y-4)²=37

Let the center of the circle (m,n) then: (m-4) 2+(n+2) 2=(m+1) 2+(n-3) 2

Yes: 16-8m+4+4n=2m+1+9-6n so: m=n+1

Circle c: (x-n-1) 2+(y-n) 2=(n-3) 2+(n+2) 2

The length of the line segment on the y-axis is 4 number 3: (n+1) 2+(y-n) 2=(n-3) 2+(n+2) 2

y^2-2ny-(12-4n)=0 so yi+y2=2n; y1*y2=-(12-4n)

y1-y2)^2=(y1+y2)^2-4y1*y2=4n^2+4(12-4n)=48

That is: n 2-4 n=0 gives n=4 or 0

If the radius is less than 5, delete n=4, so n=0

Circle c: (x-1) 2+(y) 2=13

Hint: Let the general equation of the circle, substitute p and q, and subtract d and f, so that x=0: to get a quadratic equation about y, and find e under the root number [(y1+y2)squared-4y1y2] using Vinda's theorem, chord length = 4 roots and 3 = roots

1) Solution: Let the center of the circle a(a,b).

then (a-4) 2+(b+2) 2=(a+1) 2+(b-3) 2 simplification yields b=a-1 then a(a,a-1).

a 2+(2 root number 3) 2=(a-4) 2+(a-1+2) 2 gives a = 1 or 5

Because the radius is 5

So the square of the radius of a=1 a(1,0) is (1-4) 2+(0+2) 2=13

The circular equation is (x-1) 2+y 2=13

2) Solution: What is the O point?

The wrong question should be acb=90°, and we did it today......No.

The center of the circle is on the perpendicular bisector of PQ, the slope of the straight line PQ K1=(3+2) (1-4)=-1, the coordinates of the midpoint of the PQ m, MX=(4-1) 2=3 2, my=(3-2) 2=1 2, M(3 2, 1 2), the slope of the vertical bisector of the line segment PQ is its negative reciprocal, K2=1, the equation: (y-1 2) (x-3 2)=1, y=x-1, let the coordinates of the center of the circle be c(a, a-1), then the circle equation is: (x-a) 2+(y-a+1) 2=r 2,(1)

r is the radius of the circle, let the circle and the y-axis intersect at m and n points, take the midpoint e e of mn, connect ce, then ce mn, and |ne|=|mn|/2=2√3，|ce|The abscissa of the center of the circle is a, according to the Pythagorean theorem, r 2=a 2+(2 3) 2, substituting (1), (x-a) 2+(y-a+1) 2=a 2+12, (2).

P point coordinates are substituted into equation (2), a 2-6a+5=0, a = 1, a = 5, r 2 = 13, or r 2 = 37, the circle equation is: (x-1) 2 + y 2 = 13, or (x-5) 2 + (y-4) 2 = 37

The center of the circle must be on the perpendicular line in the PQ.

pq midpoint (3 2, 1 2) pq slope:

PQ Perpendicular:

y=x-1, let the center of the circle (t, t-1).

The radius of the circle is squared.

t-4) 2+(t+1) 2=2t 2-6t+17 The distance from the center of the circle to the y-axis is.

t|∴|t|2+(2 3) 2=2t 2-6t+17 t 2-6t+5=(t-1)(t-5)=0 t=1 or t=5

t=1, the center of the circle (1,0), and the circle equation is.

x-1)^2+y^2=13

t=5, the center of the circle (5,4), and the circle equation is.

x-5)^2+(y-4)^2=37