Straight line equation for a circle! Ask for a detailed explanation, and you can get extra points if

then |p1q|、|p2q|The maximum and minimum values of the distance from point p to the straight line 3x+4y+12=0 are respectively.

cq|=|-6+12|/√(3²+4²)=6/5

p1q|=|cq|+r=11/5

p2q|=|cq|-r=1/5

Then y-2=k(x-1), i.e., kx-y+2-k=0

The straight line y-2=k(x-1) passes the fixed point (1,2).

The crossing point (1,2) is the two tangents of the circle l1 and l2

The distance from the center of the circle to the straight line kx-y+2-k=0 is.

d=|-2+2-k|/√(1+k²)

The circle is tangent to the straight line.

d=r|-2+2-k|/√(1+k²) = 1

Simplify, obtain: 8k -12k + 3 = 0

The solution is k=(3 3) 4

k(max)=(3+√3)/4

k(min)=(3-√3)/4

That is, the maximum and minimum values of (y-2) (x-1) are (3+ 3) 4 and (3-3) 4, respectively.

Method 2: (Troublesome, easy to understand).

You can also have simultaneous equations.

kx-y+2-k=0

x+2)²+y²=1

Get: (1+k)x-2(k-2k-2)x+(k-4k+7)=0

2(k²-2k-2)]²4(1+k²)(k²-4k+7)≥0

Simplify, obtain: 8k -12k + 3 = 0

Solution (3- 3) 4 k (3+ 3) 4

k(max)=(3+√3)/4

k(min)=(3-√3)/4

That is, the maximum and minimum values of (y-2) (x-1) are (3+ 3) 4 and (3-3) 4, respectively.

The picture is here: (look at the bottom, right-click to save as).

1) The distance from the circle to the straight line is 6 5 rules.

Maximum: 6 5 + 1 = 11 5

Minimum: 6 5-1 = 1 5

2) Let q(1,2) find the range of kqp.

Let l:y=kx-k+2 be the tangent of the circle, then k=(3+ 3) 4 or (3-3) 4

qm:y=[b (a-1)]*x-1), so that x=3, y=4b (a+1), then a(3,4b (a+1)).

pm:y=[b (a+1)]*x+1), so that x=3, y=2b (a-1), then b(3,2b (a-1)).

Circle c with AB as diameter: (x-3) 2+(y-4b (a+1))(y-2b (a-1)) = 0(1).

The above steps are very natural, and the difficulty lies in how to find the fixed point below.

If you combine the terms directly, it will be cumbersome, and there are two unknowns, so to simplify the operation, you can use the parametric equation here.

Let a=cos ,b=sin ,4b (a+1)=4sin (cos +1)=(8sin 2*cos 2) (2cos 2)=4tan 2,2b (a-1)=2sin (cos -1) = (4sin 2*cos 2) (2sin 2) = -2cot 2 (double angle formula).

1) becomes: (x-3) 2+(y-4tan 2)(y+2cot 2)=0

To eliminate, just make y=0 and (1) becomes: x 2-6x+1=0

Solution: x=3+2 2,3-2 2

Thus: (x,y)=(0,3+2 2), (0,3-2 2) is always the two solutions of (1).

Therefore, the circle c is constant (x,y)=(0,3+2 2),(0,3-2 2).

Here the value of y is assigned, the variable is eliminated, and then the corresponding x is solved, which is a common method to solve the problem of constant fixed point, and the correctness of the calculation is also illustrated by the drawing software (as shown in the figure, the three circles obtained by m at different positions are all over the fixed point).

The equation for the straight line is not mx+2ny-4=0 The straight line must pass through the center of the circle (2,1) so 2m+2n=4

So m=2-n mn=2n-n2 is less than or equal to 1

Bisect the circumference of the circle x 2+y 2-4x-2y-4=0.

The straight line mx+2ny=4 is over the center of the circle.

And the center of the circle (2, 1).

So. 2m+2n=4

n=2-mmn=m(2-m)=-m-1) 2+1 The value range of 1mn is (- 1].

The meaning of the quantity product of the vector is used, and the vector oa vector ob=|oa|×|ob|cos, where is the angle between the vector oa and the vector ob, due to |oa|=|ob|=1, then there is cos =a. Note that this is the positional relationship between a straight line and a circle, and that the perpendicular diameter theorem is the most common tool for this type of problem. If you do od ab through the center of the circle, the vertical foot is d, then cos aod=od oa=od, note =2 aod, then cos =2cos 2( aod) 1, that is, a=2(od) 2 1, where od is the distance from the point o to the straight line x y=a, which is |a|2, substitution to get a = a 2 2 1, that is, a 2 2a 1 = 0, you can solve the value of a.

y=a-x substituting x 2+y 2=1 gets 2x 2-2ax+a 2 -1=0 to get x1x2= (a 2 -1) 2, x1+x2=a, y1y2= a 2 -a(x1+x2)+x1x2= (a 2 -1) 2,, substituting x1x2+y1y2=a, a 2-a+1=0,, a=(1 plus minus 5) under the root divided by 2

It's very simple. After connecting the equation of a straight line and the equation of a circle, the elimination element, which is a quadratic equation about x or about y, and then use Vedd's theorem (the relationship between roots and coefficients) to do it. The one inside is changed.

Typo is wrong. Sleepy. Excuse me.

I don't fight again.

Let the equation of the line y-2=k(x+1), i.e., kx-y+2+k=0 be tangent to the circle x +y =5, that is, the distance from the origin is 5

Using the formula for the distance from a point to a line, 5=|2+k|(k +1), 4k -4k + 1 = 0, solution k = 1 2

So the equation for a straight line is 1 2x-y+5 2=0.

Actually, why is there only one solution? It's because the point is on the circle.

In fact, it is only necessary to use one theorem: the tangent equation for a point (a, b) over x +y = c (constant) is ax + by=c.

Let y=kx+k+2

Then bring y into x +y = 5 so that the equation about x has only one root. That's it, k=

Let the straight line be y=kx+b, then.

2=-k+b

The square of b 2=5 (the formula for the distance from the point to the line, squared on both sides) to find the value of k b, and substitute it into the straight line formula.

2. Five forms of linear equations:

Name: Form of Equation, Geometric Meaning of Constant, Scope of Application.

The point oblique y-y1=k(x-x1) (x1,y1) is a fixed point on a straight line, and k exists.

A straight line that is not perpendicular to the x-axis.

Oblique truncation y= kx+b k is the slope and b is the intercept of the line on the y-axis.

A straight line that is not perpendicular to the x-axis.

Two-point y-y1y2-y1 = x-x1x2-x1

x1≠x2,y1≠y2

x1,y1), x2,y2) are two fixed points on a straight line, not perpendicular to the x-axis and y-axis.

Intercept xa+yb =1

a,b≠0) a is the non-zero intercept of the line on the x-axis, b is the non-zero intercept of the line on the y-axis that is not perpendicular to the x-axis and y-axis, and is not the origin.

The general formula ax+by+c=0

a2+b2≠0) has a slope of -ab and an intercept of -ca on the x-axis and -cb on the y-axis

Straight lines in any position.

1. Definition of a circle, standard equation, (x-a)2+(y-b)2= r2; Parametric Equation:

2. The general equation of the circle: x2+y2+dx+ey+f=0 The formula has a circle center (-d2, -e2) with a radius of 12d2+e2-4f; It reflects its algebraic characteristics: the x2+y2 coefficients are the same and all are 1, excluding the xy terms.

Let's talk about the idea first, if we want to make the line and the circle have an intersection point, that is, the distance from the center of the circle to the straight line is less than or equal to the radius, we can follow this equation, it is easy to know the coordinates of the center of the circle (a, 0), the radius is 2, and the distance formula from the point to the line is |ax+by+c|a 2 + b 2, this question a=1b=-1c=1,x=a,y=0, so|a+1|2< = 2, solution -3<=a<=1

Straight line x-y+1=0

i.e. y=x+1

Substituting (x-a) 2+y 2=2

Get (x-a) +x+1) =2

Simplification of 2x -(2a+2)x+a -1=0 because lines and circles have a common point.

So =(2a+2) -4 2(a -1) 0 is simplified to -1 a 3

Center coordinates (a,0) radius r= 2

If there is a common point between the line x-y+1-0 and the squared of (x-a) + the square of y=2, then the distance from the center of the circle to the line d<=r

d=|a+1|/√2<=√2

a+1|<=2

3<=a<=1

Radius = (3, -2) distance from straight line = |3+4+1|/√1²+(2)²=8/√5

So the equation for the circle is:

x-3)²+y+2)²=64/5

The circle is tangent to the straight line, then the distance from the center of the circle to the straight line is the radius, and only the distance from him is required. The final answer is (x-3) 2+(y+2) 2=64 5