High school math problem Solve the equation part of the 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 +

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.

If we know that there are two points a(a,b)b(m,n), then the circle equation with ab as the diameter is (x-a)(x-m)+(y-b)(y-n)=0.

1.Solution: The standard equation of the circle is (x-2) +y+3) =25, so the center of the circle o is (2, -3), connecting the center of the circle and the point a knows that the line oa is perpendicular to the chord, and the slope of oa k1=[-2-(-3)] (4-2)=1 2, according to the vertical, k*k1=-1

The slope of the string is k=-2, and the linear equation can be obtained by passing the point a through the string oblique formula: y-(-2)=-2(x-4).

Simplification: 2x+y-6=0

2.Solution: Do it according to the geometric relationship.

The standard equation for a circle is (x-1) +y-1) =1, so the center o coordinate of the circle is (1,1) and the radius is 1

The distance from the center of the circle o to the straight line l is d=l3 1+4 1+8l (3 +4 )= 3

Drawing according to the geometric meaning can be seen:

The minimum value of the distance is d - radius 1 = 2

The maximum value of the distance is d + radius 1 = 4

3. Solution: If the slope of the straight line l does not exist, due to the crossing point (-5, -10).

So the equation for a straight line is x=-5

At this point, the straight line is tangent to the circle and has no chord length. So the slope of the straight line exists.

Let the slope of the line l be k, then the equation is y-(-10)=k【x-(-5)].

That is, kx-y+5k-10=0

According to the equation of the circle, the center of the circle is (0,0) and the radius is 5

The distance from the center of the circle o to the straight line l is :

d = [5 - (5 2 2)] under the root number = 5 2 2

From the formula for the distance from the point to the line: the distance from o to l is.

d=5√2/2=l5k-10l/√（k²+1）

The solution yields k = 1 or k = 7

So the linear l equation is x-y-5 = 0 or 7x-y + 25 = 0

4. Solution: Because the chord length of the straight line i is truncated by the circle c is 2 3 according to the geometric relationship.

The distance from the center of the circle to the chord is d = [2 -(3)] = 1 under the root number

According to the equation of the circle c, the center of the circle c: (a, 2) and the radius r=2

According to the distance from the center c of the circle to the straight line l:x-y+3=0 is 1, the distance formula from the point to the straight line is applied

d=1=la-2+3l/√2

The solution yields a=-1+2 or a=-1-2

1.Circle center (2,-3),a(4,-2) k1=-2-(-3) 4-2=1 2,k*k1=-1 k=-2, point oblique can be obtained.

y=sint+1, d=absolute value [3cost+4sint+15] 5, the maximum value is 4, and the minimum value is 2

The chord length is 5 2, d=5 2 2, [5k-10] (k 2+1)=d, and k is a straight line.

4.In the same way, thirdly, d=1, we can find a [] as the absolute value.

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.

x^2+y^2-6mx-2(m-1)y+10m^2-2m-24=0,x-3m)^2+(y-(m-1))^2=25

The coordinates of the center of the circle are (3m,m-1), and the radius of the cherry blossom lead 5 is set to the line l, and the line parallel to the line l is set to l':

x-3y+n=0

The center of the circle is rented to the straight line L'The distance d=|n+3|10 has nothing to do with m.

Using the Pythagorean theorem :)

The chord length 2 ridge is good (25-d 2) and has nothing to do with m.