Sophomore Math Circle Questions! Senior 1 math circle problem

Two common ideas, one is the commutation method, and the other is the combination of numbers and shapes, using slope!

Here is the answer with the idea of changing the yuan:

The point m(x,y) is the moving point on the circle x 2+y 2=2y, and the equation for the circle can be formulated into the standard equation: x 2 +(y-1) 2=1 can make x=cos , y-1=sin

Then 2x+y= (2cos)+1+sin )=(2cos +sin)+1= 5cos( +a)+1, where a is an auxiliary angle, and since -1 cos( +a) 1, the range of 2x+y is [1- 5,1+ 5]. I wish you to study hard and make progress every day!

Solution: Let 2x+y=k, substitute x y -2y=0, and get x +(k-2x) -2(k-2x)=0

i.e. 5x +4(1-k)x+k -2k=0 discriminant = 16(1-k) -20(k -2k) 0

1-root number 5 k 1 + root number 5 is the desired range.

The circular equation is x 2 (y-1) 2 1

Let x=sin a , y=1+cos a , 0<=a<360 degrees.

2x+y=2sin a +1+cos a

There is an auxiliary angle formula that knows that the maximum value is 1 + root number 5, and the minimum value is 1 - root number 5

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.

Hello first question.

a=c, if a is not equal to c, then it cannot be divided by a number, and it is reduced to the general form (x-a) 2+(y-b) 2=r 2 (both x and y have a coefficient of 1).

Further, if it is equal to 0, then the x 2 term and y 2 terms do not exist, and there is no equation for a circle.

The second problem, x-a) 2+(y-b) 2=r 2 This is the equation for a circle, and then, without the xy term, it means that b=0, and the xy term is only =0

The third question, d 2+e 2-4af, you should have made a mistake, is d 2+e 2-4f

d 2 + e 2-4f is the radius of the circle, which can be obtained by the formula in the form of (x-a) 2+(y-b) 2=r 2.

Try your own recipe and convert the equation of the circle into (x-a) 2+(y-b) 2=r 2 to solve the problem at the same time: "Let the equation of the circle be x 2+y 2+dx+ey+f=0", then the center of the circle is -(d 2), -e 2)".

What does it mean, you can hi me if you don't understand.

There is a common way to solve the problem of circles, the formula, which is formulated as (x-a) 2+(y-b) 2=r 2, so that all problems can be solved. The size and position of the two circles were determined, and the drawings were drawn, and they were all done.

As I said upstairs, it's very simple, knowing the center of the circle and the radius, what else can't be solved, the second question: simultaneous equations, eliminate x 2 + y 2 and get the straight line equation directly. The third question is to find the distance from this point to the center of the two circles, and then set the point (x,y) to solve the equation.

1. Suppose the center of the circle is (m,n), because the points (4,-1) and points (9,6) are both points on the circle, so the distance between the center of the circle and these two points is equal, that is, (m-9) +n-6) =(m-4) +n+1), and 50-5m=7n,.. is obtainedType 1.

And because the circle and the straight line x-6y-10=0 are tangent to (4,-1), the equation for the diameter of the circle through (4,-1) is y+1=-6(x-4).

So the center of the circle (m,n) is written in the equation of Eq. 2 as n+1=-6(m-4)).Type 2.

It is obtained from equation 1 and equation 2, m=3, n=5

That is, the center of the circle is (3,5).

The radius of the circle is 37

So the equation for a circle is (x-3) +y-5) =37

2,x +y -2x + 4y = 0, i.e.

x-1）²+y+2）²=5

Indicates a circle with the center of the circle at (1,-2) and a radius of root number 5.

Let x-2y=b, which represents a system of straight lines, which varies with the value of b.

The maximum value of x-2y satisfying x +y -2x + 4y = 0 is the maximum value of b when a circle and a straight line system intersect.

You can draw the diagram below, and it is easy to see that there is a maximum value when the line and the circle are tangent (the top is the maximum, and the bottom one is the minimum).

The distance from the center of the circle (1,-2) to the line x-2y=b is equal to the radius of the circle 5:

5-b|/√5=√5

5-b|=5

b = 10 or b = 0

b=10 is the maximum value sought, and b=0 is the minimum value.

Using the idea of combining numbers and shapes, we draw an image of a circle.

Let x 2 + y 2 = r 2, which is also regarded as a circle, which is equivalent to asking where the point is in a known circle, so that the line between the center of the circle and the point is minimized.

That is, the origin and the center of the circle can be connected, and the minimum value is equal to the length of the line between the center of the circle and the origin minus the radius.

As shown in the figure below. <>

That is, to find the minimum value from the point on the circle to the origin, you can connect the origin and the center of the circle, and the minimum value is equal to the length of the line between the center of the circle and the origin minus the radius.