High school circle problem, high school circle maximum value problem induction

Note that the slope of the straight line l does not exist first, that is, x=-3, and the chord length in this case is also 8, which is in line with the topic.

Then consider the existence of the slope of the straight line l, let the slope be k, the straight line equation is y+3 2=k(x+3), and then find the distance from the center of the circle (0,0) to the line l of the set, half of the length of the chord and the distance and radius from the center of the circle (0,0) to the line l can form a right triangle, using the Pythagorean theorem, r = 5, the length of the chord is 8, so the distance from the center of the circle o to the line l is 3. ∴|3k-3/2|(k2+1)=3, and k=-3 4Then you can get it by putting k = -3 4 generations in.

To sum up the above two situations, there are two solutions to the straight line l: x=-3 3x+4y+15=0, if you haven't understood it, tell me the mailbox, and I'll write the solution process on paper and send it to you.

Take a look at the diagram and calculate it with the vertical warp force.

Let the equation of the straight line, p(-3,-3 2), y+3 2=k(x+3) and because r=5 and the chord length is 8, the distance from the center of the circle o to the straight line l is 4o(0,0), r=5, kx-y+3 2=0(3 2) (k2+1)=r=3, and k=0l=-3 2

I suspect you made a mistake).

The maximum value problem of the high school circle is summarized as follows:

Type 1: "The maximum value of the distance from a point on a circle to a straight line" problem.

1. Find the maximum and minimum distances from the point on the circle c:(x-2) +y+3) =4 to the straight line l:x-y+2=0. Analysis: As chii intersects H, intersects with circle C at A, and inversely extends with circle at point B.

So d=7=-7+2-d7-2.

2. Find the maximum and minimum values of the distance between the point on the circle c: (x-1) +y+1) =2 and the straight line l:x-y+4=0.

Analysis: The method is the same as the first question, d=dm=3 2+ 2=4 2; d=3√2-√2=2√2

3. The minimum value of the distance from the point on the circle x2+y2=2 to the straight line l: 3x+4y+25=0 is.

Analysis: The method is the same as the first question, d....=5-√2

Type 2: "The maximum value of the distance from a point on a circle to a fixed point" problem.

1.It is known that the point p(xy) is the point above the circle c:x+y-2x-4y+4=0, and the maximum and minimum distance from the imaginary slip p to the original difference Huila point is obtained.

Analysis: Connect OC and circle intersection at A, extend OC intersection with B dmax=doc+r=√5+1; dmin=doc-r=√5-1.

The answer to the first question is (x+1) 2+(y-2) 2=20, and the answer to the second question is yes.

3x-4y+6=o

The third question is that I don't know how to do it. Does the third question mean that the result of the second question can be used for the third question?

Solution: Let the center of the circle o

1]: a(-1,2) is the tangent of the circle centered on the line l1:x+2y+7=0 and the distance from a to l1 is the radius r

The absolute value of r=(-1+4+7) 5= 20 The equation for a circle is (x+1) +y-2) =20 mn=2 19

i.e. qm or qn is 19

And q is the midpoint of Mn.

i.e. OQ vertical MN

oq=√(20-19)=1

Let l be y=kx+b

k≠0) and its pass b(-2,0).

can be changed to y=kx+2k

and the distance from o to l is 1

i.e. the absolute value of (2-k) (k 1=1

Get k for 4 3

The solution is 3x-4y+6=0

3] Vector bq vector bp = modulo multiplication vector of vector bq modulo multiplication vector bp cos angle and its collinear.

That is, the value is the length of the vector bq multiplied by the length of the vector bp.

And its value is positive.

Consider setting t=(the length of the vector bq multiplied by the length of the vector bp) and l=y=kx+2k

oq²=（k²-4k+4)／(k²+1)

The distance from a point to a straight line can be found).

And ob = 5

qb =ob -oq =(4k +4k+1) (k +1) by the intersection of the line l and l1 at the point p, which can be expressed as x=(-7-4k) (2k+1).

y=(-10k)／(2k+1)

then bp can be expressed as (25k +25) (k -4k+4) t=bp qb =25

The vector bq vector bp is a fixed value.

You can draw a picture of the problem with a fixed value of 5, and then combine it with my ideas, which should be fine.

Adopt me, I did it very carefully. Didn't find any problems. Thank you.

It can be proved by set theory, the equation x 2+y 2+d1x+e1y+f1+ (ax+by+c)=0 satisfies the general equation of the circle, so this equation describes a circle, and all the points (i.e. intersections) that meet x 2+y 2+d1x+e1y+f1=0 and ax+by+c=0 must satisfy x 2+y 2+d1x+e1y+f1+ (ax+by+c)=0, because 0+ *0=0, so, Their intersection is on the circle determined by this equation (belonging to the set described by this equation). However, for points that do not satisfy x 2 + y 2 + d1x + e1y + f1 = 0 and ax + by+ c = 0, x 2 + y 2 + d1x + e1y + f1+ (ax + by+ c) = 0 are the other points on this circle that are not two intersections. For example, x 2+y 2+d1x+e1y+f1+ (ax+by+c) 2=0 , this equation describes an ellipse (including imaginary ellipses) that passes through the intersection of a straight line and a circle.

For any number of systems of equations, each of which contains a number of equations, most of their intersection spaces can be constructed and contained in the space that satisfies the conditional features.

Just remember, the exam won't test the derivation process, kid.

Since the area of PCD is 12, we know that the point p can only be on the straight line: y=x+10 or y=x-2, and we know that there are only 3 such points, then we can judge that E is tangent to the straight line y=x+10 (only one p point satisfies), and y=x-2 intersects (two p points satisfy), it is obvious that the coordinates of e can be expressed as (a 2, a 2), the distance from point e to the straight line y=x+10 is equal to the distance from it to any point of a, b, o, there is only one unknown number a, find a can be found, Do the specific calculations yourself, and ask me if there is anything you don't understand.

Solution: On the circle x +y = 50, there are 12 integer points, (1, 7), (5, 5), (7, 1).Any two points can be used as a straight line, which can be made as c(12,2)=66, plus 12 tangent lines, a total of 78. Therefore, D. was chosen

With (2,0) as the center of the circle and the root number 3 as the radius as the circle, the y ratio x can be regarded as the slope of a straight line passing through the origin, and when the line is tangent to the circle, it reaches the maximum or minimum value.

Let y x k find the slope of the maximum value of k.

The line is tangent to the circle, and the distance from the center of the circle to the line y=kx = 3

You can find the maximum value of 1

A circle with (2,0) as the center of the circle and root number 3 as the radius.

The maximum value of y x is the slope of the tangent = 3 + 2 root number 3

When the center of the circle is exactly on the straight line x-2y=0, then I personally think that this circle is what we want.

d=|Root No. 3 3-2b|Root number 5

Only when the root number 3 3-2b = 0 is in line with the title, that is, b = root number 3 6.

The radius r=2 root number 3 3, and the circle equation is (x-root number3 3) 2+(y-root number3 6) 2=2 3

Well, by the title, it seems that the equation for the circle given in the title should be of circle 1?

If it is a circle 1, then the equation for determining circle one can be rewritten as: (x-1)2+(y+2)2=5-m, so that the coordinates of the center of the first circle can be found, which is (1,-2). And we already know the coordinates of the center of the second circle.

According to the meaning of the title, the straight line can be set to y=kx, so that the distance from the line to the center of the two circles is equal to the radius of the two circles, but it seems that m is not used? Should there be any other conditions?