Senior 1 Mathematics In addition to the equation that connects the circle is Please answer in detail

l1:x+3y-12=0,l2:3tx-2y-2=0 The quadrilateral has a right-angled vertex - the coordinate origin, and the other two vertices on the coordinate axis are connected by the diameter, and the vertex that is not on the coordinate axis - the fourth vertex must also be a right-angled vertex, so the two straight lines are required to be perpendicular to each other, that is, the slope is negative to each other

1 3=-1 (3t 2), so l2:3x-y-1=0, respectively, let x=0, y=0, and get the intersection of l1, l2 and the coordinate axis is:

a(0,4),b(12,0),c(0,-1),d(1 3,0)l1,l2 are combined to find their intersection points: e(3 2,7 2) so the quadrilateral vertices are o(0,0),a(0,4),e(3 2,7 2),d(1 3,0),ad|=(1 3-0) +0-4) =145 9, so the radius squared is (|.)ad|2) =145 36, the midpoint of AD is the center of the circle, and its coordinates are (1 6, 2).

So the equation for the garden is: (x-1 6) +y-2) =145 36

Because the diagonal complementarity of the quadrilateral inside the circle should be perpendicular to each other, i.e., 3t 2=3, giving t = 2. The straight line l2 is 3x-y-1=0, the intersection point of the two lines is (3 2,7 2), and the two intersection points of the positive half axis of the coordinate axis are (0,4) and (1 3,0) respectively. So, the center of the circumscribed circle is (1 6,2), the radius is 145 6, and the equation for the circle is (x-1 6) +y-2) =145 36

Solution: The center of the circle c:(x-1) 2+(y-2) 2=5-m is (1,2).

The distance to the straight line l:x+2y-4=0 is:

d=|1+2*2-4|/√5=√5/5

Strings|mn|=4/√5=4√5/5

So r 2 = ( 5 5) 2 + (4 5 5) 2 = 17 5 i.e. 5-m = 17 5

So m=8 5

In high school math, when it comes to straight lines and circles, you must be able to do this by yourself.

The conic curve at the back is the most important.

And it's a must-have exam every year.

Practice a lot on your own.

The process of solving the problem is as follows:

Circle C formulation.

x - 1) 2 + y - 2 ) 2 = 5 - m The center of the circle is (1,2).

Uses the formula for the distance from any point to a straight line.

Find the distance from the center of the circle to the straight line l:x+2y-4=0 as:

d =|1 + 2*2 - 4|5 = 5 5 stringsmn|=4/√5 = 4√5 / 5

So the Pythagorean theorem is applied in right triangles.

r 2 = ( 5 5) 2 + 4 5 5) 2 i.e. 5 - m = 17 5

m = 8 / 5

x 2+y 2 is the square of the distance from the point on the circle to the origin.

The distance from the center of the circle (-1,1) to the origin is 2, and the radius of the circle is 2, so the maximum value of x 2+y 2 is 2 8 squared

Let the circle of a, b, and c be x 2 + y 2 + dx + ey+ f = 0 then -d + 5e + f + 26 = 0

5d+5e+f+50=0

6d-2e+f+40=0

The solution yields d=-4, e=-2, f=-20

The equation for a circle is x 2 + y 2-4x-2y-20 = 0 substituting (-2, -1) the above equation is not true, and point d is not on the circle.

Such a circle does not exist.

3 points to determine a circle, set the general formula to solve d, e, f. Then we can determine whether the remaining point is on the circle or not.

There is no intersection. The distance from the straight line x0x+y0y=r 2 to the center of the circle o(0,0) is d=r 2 (x0 2+y0 2), and since p(x0,y0) is the inner point of the circle, (x0 2+y0 2) has no focus and is separated from the circle.

Question 1: The equation can be set to x 2 + y 2 + ax + by + c = 0

1,2), (3,4) substitution:

a+2b+c+5=0 ……1）

3a+4b+c+25=0 ……2）

The chord length on the x-axis is 6, that is, the difference between the two roots of the quadratic equation about x is 6 when y=0, so that y=0:

x^2+ax+c=0

The relationship between the root and the coefficient: x1+x2=-a, x1x2=c, has: |x1-x2|=6

x1-x2|=√x1+x2)^2-4x1x2]=√a^2-4c)=6

i.e.: a 2-4c = 36 ......3）

Solve the system of ternary quadratic equations composed of (1), (2), and (3) to obtain:

a1=12，b1=-22，c1=27；

a2=-8，b2=-2，c2=7。

The equation for substituting the circle is as:

x 2 + y 2 + 12x-22y + 27 = 0, or x 2 + y 2 - 8x - 2y + 7 = 0

i.e., (x+6) 2+(y-11) 2=130, or (x-4) 2+(y-1) 2=10

Question 2: Introduce circular equations.

Let the circle be: x 2 + y 2 + 6 x - 4 + k (x 2 + y 2 + 6 y - 28) = 0....1)

x^2+y^2+6/（1+k)x+6k/(1+k)y+(-4-28k)/（1+k)=0

So the center of the circle is x=-3 (1+k) y=-3k (1+k).

Substituting x-y-4=0

Solve k = -7 and bring in (1) to get the equation obtained.

Give it points.