A problem that has been difficult for me for several days in the second compulsory math of the first

The value of <> is the maximum value of the slope of the moving line op.

The center of the circle is a(3, root number 3), radius r = root number 6, p point changes on the circle, it is easy to see that the position as shown in the figure is the place where the op slope obtains the maximum value, that is, the tangent point (the other tangent point is in the fourth quadrant, which is not drawn in the figure, but it is obvious that the op slope is negative when the tangent point is taken, so it is not considered), let y=kx, and connect it with the equation of the circle to obtain an equation containing only one variable x, and then make its discriminant formula b 2-4ac=0, (that is, the line and the circle have only one intersection point), Find the coordinates of the intersection point, round off the tangent coordinates of the fourth quadrant, and obtain the coordinates of the p point, then the maximum value of y x is easy to obtain.

This problem can be regarded as finding the maximum value of (y-0) (x-0), which is similar to finding the slope of a straight line, that is, finding the straight line with the largest slope of the point and the origin of the circle at the same time, then this problem can be transformed into finding the slope of the tangent of the circle that has passed the origin, but this will find two values, whichever is larger.

In fact, it is very simple, the maximum value of y x=tana is equal to the value of po (o is the origin of the coordinate system) and the tangent of the circle;

The three sides of the right triangle at this time: root number 6,? ，3

Pythagorean theorem:? *=3*3-6=3

Root number 3tana = root number 6 Root number 3 = root number 2

The maximum value of y x is equal to root number 2

Circle x 3 y 3 6 Circle center 3, 3 Radius 6 6 Circle center 3, 3 The greater value of the distance 12 z from the center of the circle is: Radius Center 3, 3 distance from the dot At this time, the circle o is inscribed with the circle with the origin of the radius z Z 2 3 6

To make y x maximum, it is the slope of the straight line tangent to the circle over the origin.

Or another y x=k to connect it with the equation of the circle is the maximum value of Dell's solution of k greater than or equal to 0.

Let x= 6sina+3;y=√6cosa+√3

Then y x=(6cosa+ 3) (6sina+3)==(2cosa+1) (2sina+ 3).

Simplify it yourself, and that's how it should be.

1. After the fixed point (3,1), the straight line becomes the form of am+b=0, a=0, b=0, and if it is established at the same time, you can solve x = 3 and y = 1.

2. You find the coordinates of the center of the circle, and then calculate the distance from the center of the circle to the straight line, compare it with the radius, and discuss it by classification;

3. Calculate the chord centroid distance first, and then calculate the chord length according to the Pythagorean theorem.

Let the straight line be x a+y b=1 through the point a(-2,3), so 3a-2b=ab and the area of the triangle enclosed by the straight line and the two coordinate axes is 4, so ab=8 or -8 joint excavation of the simple spike equation system is solved to determine a=4,b=2 or a=-4 3,b=-6, so x 4+y 2=1 or vertical x -4+y -6=1, that is, x+2y-4=0 or 3x+2y+12=0 is sought.

y - 3 = k(x+2)

x = 0, y = 3+2k,y = 0, x = 2 -3/k

and the intersection of the coordinate axis to pretend to be Wang Rangzi as missing (0, 3+2k), 2 -3 k, 0).

Triangle area: |3+2k|*|2 -3k|/2 = 4

You can set the linear equation y=kx+b, so that the intersection point with the y-axis is (0,b), and the intersection point with the x-axis is with pure (-b k,,0).

Then the area is 4, that is, the absolute value is 4 and because of the fixed point (-2,3) so 3=-2k+b

It's OK to solve it.

(x2)+(y2)-2x+4my+3(m2)=0(x-1) 2+(y+2m) 2=4m 2+1-3m 2=m 2+1 The center coordinate of the circle is (1,-2m), on the line x+y+2=0, then there is:

1-2m+2=0

Get m=3 2

Radius = root number (m 2 + 1) = root number (9 4 + 1) = root number 13 2

First, the equation of the circle is organized into a standard form, i.e., (x-1)2+(y+2m)2=m2+1

In this way, we can see that the center of the circle is (1,-2m), and substituting the coordinates of the center of the circle into the linear equation gives m=3 2

r) 2=m 2+1 solution r=root number 13 2

Chord length a = 6 2

radius r=2 5

So by the Pythagorean theorem.

The chord heartstep d= [r -(a2)]= 2

Let the straight line be y+4=k(x-6).

kx-y-4-6k=0

So d=|0-0-4-6k|(k +1) = 2 squared, 36k +48k+16=2k +2

17k²+24k+7=0

k=-1,k=-7/17

So it's x+y-2=0 and 7x+17y+26=0

The first problem finds the slopes k of ab and bc respectively, if they are the same, the three points are collinear, and if they are different, they are not collinear.

You can also use the method of our junior high school, set y=kx+b, bring in a b coordinate to find the analytic formula, and then bring in the point c abscissa to see if the ordinate is 12

In the second question, first find the coordinates of the midpoint of ab (7-5 2, -4+6 2), then the straight line perpendicular to ab must pass this point, and then find the slope k of ab, because the product of the slopes of the two straight lines perpendicular to each other is -1, you can find the slope of the straight line, and then set the point oblique formula to find it.

Friend, I suggest you take a look at the compulsory two textbooks

1) Vector ab = (4, 4) vector bc = (5, 5) so three points are collinear.

2) The midpoint of AB (1,1), the slope of AB is -5 6, so the equation for the perpendicular bisector of the line segment AB is Y-1=-6 5(X-1).

（a+1）x+y+2=0

y=-(a+1)x-2

Because it does not pass through the second quadrant, it is known by drawing.

k=-(a+1)>0

b=-2<0

then a+1<0

a<-1

Equation for l as y= (a 1) x a 2 so that l does not pass through the second quadrant if and only if.

a＋1）≥0

a 2 0a 1 Therefore, the range of values for a is a 1

The analysis yields a straight line passing through one, three, and four quadrants.

Let x=0, y=-2<0

Let y=0,x=-2 (a+1)>0 a<-1 if a+1=0,a=-1, y=-2 satisfies.

So a -1

Solution: From the meaning of the problem, the center o of the circle is (-1,0), and the radius r=2 2 is set to the line ab as y-2=k(x+1).

i.e. y=kx+k+2

Both sides of the straight line ab can have at most two points to the same distance from ab, because there are exactly three points to ab the distance is 2, so on a certain side of ab there must be only one point to ab distance of 2, ab is translated, so that the line and the circle are just tangent, the tangent point is the point to ab distance of 2. Set to point q

Connect oq to c, and find oc=oq-qc=2 2- 2= 2, so the distance from the center of the circle to ab d=|-k+k+2|(k +1) = 2 to get k = 1

Therefore, the equation for ab is y=x+3 or y=-x+1

Through o, two tangents of a circle (the center of the circle is c): ot1, ot2, cot1 = cot2 = 30°, aco = 135°, so 135°-30° aob 135° + 30°, that is, 105° aob 165°

Combined with the image, it can be seen).

If o is the center of the circle, then it is 0 to 180° (angles over 180° are considered obtuse angles less than 180°).