Solving equation problems for high school math circles

Let the equation of the straight line be y=kx+1, and it can be seen from the image that the tangent with the circle is the two maximums, and the distance from the point c to the straight line y=kx+1 is less than or equal to 1, d=|2k-3+1|(k +1)<=1, we get (4- 7) 3 k (4+ 7) 3

Method 1: (Combination of numbers and shapes) Let the linear equation be y=kx+1 and turn it into a general formula, that is, kx-y+1=0

Since the line and the circle intersect at two different points m and n, k must be between the two k values when the line and the circle are tangent, thus:

The distance === radius from the center of the circle c(2,3) to the line kx-y+1=0 is obtained:

d=|2k-3+1|1+k 2=1 gives k=4+ 7 3 or k=4- 7 3.

So the range of k is (4- 7 3, 4+ 7 3).

Method 2: (Discriminant Method) Bring y=kx+1 into the equation of the circle (x-1) 2+(y-3) 2=1 to obtain a quadratic equation about x, and then use the discriminant equation 0.

This question is not easy to solve with **!!

Method 1: Let the straight line y-1=kx, that is: y=kx+1 circle c: (x-2) +y-3) =1

Synoptic: x-2) +kx+1-3) =1

k²+1)x²-4(k+1)x+7=0

16(k+1)²-28(k²+1)＞0

3k²-8k+3＜0

4-√7)/3＜k＜(4+√7)/3

Method 2: Let the straight line y-1=kx, that is, the distance from the center of the circle to the straight line of kx-y+1=0.

d=|2k-3+1|/√(k²+1)＜1

4(k-1)²＜k²+1

3k²-8k+3＜0

4-√7)/3＜k＜(4+√7)/3

What about the question? The line and the circle intersect, and the distance from the center of the circle to the line dd=|2k-2|/√(1+k^2)<1

2k-2|< 1+k2) squared.

4k^2-8k+43k^2-8k+5=0

5/3

Simultaneous y=kx+1 and (x-2) 2+(y-3) 2=1, so that the discriminant is greater than zero.

What about the topic? It's better to attach a picture

The equation for a circle can be expressed as x2+y2-4y+1- (2x+y+4)=0

The smallest area is the smallest radius.

Simplify the equation and find the minimum value of r (the quadratic function of ).

Consider the equation for the circle to be x +y = a , p(x,y), m(x1,y1), and mn perpendicular to ab, so x = x1 .......

m(x1,y1) is the moving point on the circle, so x1 +y1 = a ,...Since the absolute value of op is equal to the absolute value of mn, x +y = y1 ,...Substituting "" gets: x + (x + y ) = a, i.e. 2 x + y = aThis is the trajectory equation for point p, which is an elliptic equation.

Let's write some thoughts:

First, find the distance from the center of the circle (1,2) to the straight line l, when m r, satisfying <=radius 5, then it can be proved that l and c must intersect.

When the distance is maximized, what is the value of m? At this point, the intersecting chord length is the shortest.

Then find the coordinates of the two intersection points and the distance between the two points.

Let the equation for the circle be x2 y2 4x 2y 4 (x y 4) 0 2x2 (2 2 ) x 4 4 0 let δ 0 (2 2 )2 4 2 (4 4) 0, and the solution is: 3, so the equation for the circle is x2 y2 7x y 8 0.

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.

Let's let the center of the circle c c=(a,0)a>0 and the radius r.

x²+y²-2x=0

The center of the circle (1,0) has a radius of 1|a-1|=1+r

a 2 = ra = 0 (rounded), a = 4, r = 2

In summary, the equation for the circle is: (x-4) +y = 4 correct answer, remember to adopt it*

First, we get the center o of the circle m, and the radius r

Then the distance from the center of the circle c to o minus r is the radius of the circle c, r is calculated by setting the center c of the circle to be (a,0) and the distance from the straight line is equal to r, and a circle c can be obtained.

。Here's a formula for you.

Circle 1; x 2 + y 2 + d1x + e1y + f1 = 0 round 2; x 2 + y 2 + d2x + e2y + f2 = 0 The equation of the straight line where the common chord is located (d1-d2) x + (e1-e2) y + (f1-f2) = 0

I'm so lazy... 4x+3y+13=0 Yuan, Kongqing, 1; Wang grip x 2 + y 2-9 = 0

Circle 2; x^2+y^2+8x+6y+17=0.。。Are you sure you wrote the correct equation for your circle?

1) The center of the circle is at the midpoint of the diameter ab, using the coordinate formula for the midpoint of the line segment.

The center c coordinate of the circle is [ (1+3) 2, (4+2) 2 ] = (1, 3).

The radius length is ca = [1 +1) 3-4) ] = 5

The standard equation for a circle is (x - 1) y - 3) = 5

2) Find the equation of the perpendicular bisector of the straight line ab (the center c must be on the perpendicular bisector of the chord).

Its slope k = -1 kab = -1 [(-2-1) (2-1)] = 1 3 (two lines perpendicular to each other).

Using the coordinate formula of the midpoint of the line segment, the midpoint of the line segment AB is obtained: [ 1+2) 2, (1-2) 2 ] = (3 2, -1 2).

The perpendicular bisector equation for the straight line ab: y + 1 2 = (1 3) * x - 3 2) (using the point slope).

Reduced to x - 3y - 3 = 0

And the center c of the circle is also in the line x - y + 1 = 0

The simultaneous equations x - 3y - 3 = 0 and x - y + 1 = 0 are solved to obtain the coordinates of the center of the circle (-3, -2).

The radius length is ca = [3 -1) 2-1) ] = 5

The standard equation for a circle is (x + 3) y + 2) = 25

3) The distance from the center c of the circle to the tangent is the radius length, using the formula for the distance from the point to the line.

r = 3*1 - 4*3 - 7 [3 +4)] = 16 5

The standard equation for a circle is (x - 1) y - 3) = 256 25

4) The center of the circle is on the straight line y = 2x.

Let the coordinates of the center of the circle be (a, 2a).

And the distance from the center c to the tangent is the radius length, using the formula for the distance from the point to the line (method is the same as (3),).

r = 3*a + 4*2a - 7 (3 +4 ) = 3*a + 4*2a + 3 (3 +4).

Simplification: 11a - 7 = 11a + 3

Get 11a - 7 = 11a + 3 (incompatible) or 11a - 7 = - (11a + 3).

Solution a = 2 11

From the above equation, r = 1

Get the center c coordinates (2 11, 4 11).

The standard equation for a circle is (x - 2 11) y - 4 11) = 1

1、(x-1)^2 + y-3)^2=5

2、(x+3)^2 + y+2)^2=253、(x-1)^2 + y-3)^2=

4. (x-2 11) 2 + y-4 11) 2=1 The last question can be referred to the screenshot.

1. The midpoint of the diameter is the center of the circle (1,3), where the distance from one point to the center of the circle is 5 under the root number of the radius, then the standard equation of the circle is.

x-1)^2+(y-3)^2=5

2. Let the center of the circle (x,x+1), then there is (x-1) 2+(x+1-1) 2=(x-2) 2+(x+1+2) 2 solution x=-3,x+1=-2, the center coordinate of the circle (-3,-2), and the distance to any point is the radius 5, so the standard equation is (x+3) 2+(y+2) 2=25

3. The distance from c to l is the radius 16 5, then the standard equation is (x-1) 2+(y-3) 2=256 25

4. From the question that the two straight lines are parallel, then the diameter of the circle is the distance between the two straight lines, that is, 2, then the radius is 1, the point of simultaneous y=2x and 3x+4y-7=0 is set to a(7 11,14 11), and the point of simultaneous y=2x and 3x+4y+3=0 is set to b(-3 11,-6 11), then the center of the circle is the midpoint of ab, and the coordinates of the center of the circle are (2 11,4 11), so the standard equation is (x-2 11) 2+(y-4 11) 2=1

Circular equation x 2; +(y-1) 2;=1 Let x=cos y=1+sin because x+y+a=cos +1+sin +a 0 always hold -a (cos +sin +1) the minimum value of sin +cos +