Given that the straight line L passes through the point (2,1), the inclination angle is twice of the inclination angle of the straight line x-radical 3Y + 3 = 0, the equation for solving the straight line L is obtained

Given that the straight line L passes through the point (2,1), the inclination angle is twice of the inclination angle of the straight line x-radical 3Y + 3 = 0, the equation for solving the straight line L is obtained

It is known that the slope of the straight line is root sign 3 / 3, so the inclination angle is 30 degrees
So the slope angle of the straight line is 60 degrees and the slope is root sign 3
So the equation of the straight line is Y-1 = root 3 (X-2), that is, root 3x-y + 1-2 times root 3 = 0

Basic exercises] group A
1. Given the points a (3, - 2), B (- 5,4), the equation of the circle with the diameter of line AB is 2. The equation of the circle passing through points a (1, - 1), B (- 1,1) and the center of the circle on the straight line x + Y-2 = 0 is 2
3. Given that the radius of circle C is 2, the center of circle C is on the positive half axis of X axis, and the line 0443 & ｛｛｛｛｛｛｛｛｛｛｛｛｛｛｛｛｛;, ｛｛; YX is tangent to circle C, then the equation of circle C is 4_

The middle point of 1a (3, - 2), B (- 5,4) is m (- 1,1). The middle point is the radius of the center of the circle r = am = 5. The equation of the circle is (x + 1) ^ 2 + (Y-1) ^ 2 = 252. The center of the circle must be on the vertical bisector of ab. the slope of the vertical bisector of M (0,0) at the middle point of AB k = - 1 / KAB = 1

If the point a (0,1) B (2,3) is known, the equation of the circle with the diameter of line AB is

Direct nest formula: (x-x 1) (x-x 2) + (Y-Y 1) (Y-Y 2) = 0
(x-0)(x-2)+(y-1)(y-3)=0 ,
It is reduced to (x-1) ^ 2 + (Y-2) ^ 2 = 2

If a (- 3, - 5) and B (5,1) are known, then the equation of the circle with the diameter of line AB is______ ．

It can be seen from the meaning that the midpoint of a and B is the center of the circle, so the center of the circle is: (− 3 + 52, − 5 + 12), that is, (1, - 2), the distance between AB is equal to the diameter = (− 3 − 5) 2 + (− 5 − 1) 2 = 10, the radius of the circle is 5, the equation of the circle is: (x-1) 2 + (y + 2) 2 = 25, so the answer is: (x-1) 2 + (y + 2) 2 = 25

It is known that the line L and the circle x2 + Y2 + 2x = 0 are tangent to the point t, and the hyperbola C: x2-y2 = 1 intersects at two points a and B. If t is the midpoint of the line AB, the equation% of the line L is obtained

What water? What well?
But bring back joy or sorrow, I with others
Peeling off the thick accumulation layer on the body, standing up in the clouds,
There is no need to say: finally
It's like a white wheat field
I think it's not the first try

If the line L: 2x-y + 2 = 0 is tangent to the circle (x-1) &# 178; + (Y-A) &# 178; = 1, find the value of A

A:
The line 2x-y + 2 = 0 is tangent to the circle (x-1) ^ 2 + (Y-A) ^ 2 = 1
Represents the distance from the center of the circle (1, a) to the straight line d = r = 1
So:
d=|2*1-a+2|/√(2^2+1^2)=1
So:
|4-a|=√5
So:
4-A = √ 5 or 4-A = - √ 5
So:
A = 4 - √ 5 or a = 4 + √ 5

Line L is tangent to circle X & # 178; + Y & # 178; = 2 and point P (P is not on the coordinate axis), l is tangent to hyperbola X & # 178; - Y & # 178; / 2 = 1
Intersection and different two points a, B. proving OA ⊥ ob

Let P (m, n), Mn ≠ 0
P on the circle x ^ 2 + y ^ 2 = 2, M & # 178; + n & # 178; = 2
Then the tangent of the circle passing through P is perpendicular to Op,
Slope k = - M / n
∴L:y-n=-m/n(x-m)
That is y = - M / NX + (M & # 178; + n & # 178;) / n
y=-m/nx+2/n
Simultaneous equations:
{y=(-mx+2)/n
{x²-y²/2=2
==>
x²-(2-mx)²/(2n²)=1
==>
(2n²-m²)x²+4mx-4-2n²=0
∵n²=2-m²
The equation is
(4-3m²)x²+4mx+2m²-8=0
It needs 4-3m and 178; ≠ 0
△=16m²+4(3m²-4)(2m²-8)>0
Let a (x1, Y1), B (X2, Y2)
Then X1 + x2 = - 4m / (4-3m & # 178;)
x1x2=(2m²-8)/(4-3m²)
∴OA·OB
=x1x2+y1y2
=x1x2+(2-mx1)(2-mx2)/n²
=x1x2+[4-2m(x1+x2)+m²x1x2]/n²
=[(m²+n²)x1x2-2m(x1+x2)+4]/n²
Molecule (M & # 178; + n & # 178;) x1x2-2m (x1 + x2) + 4
=2(2m²-8)/(4-3m²)+2m*4m/(4-3m²)+4
=(12m²-16)/(4-3m²)+4
=-4+4=0
∴OA·OB=0
OA ⊥ ob

If a directrix of hyperbola x ^ 2 / 16 -- y ^ 2 / K is just a tangent of circle x ^ 2 + y ^ 2 + 2x = 0, then the value of K is zero

The formula is (x + 1) ^ 2 + y ^ 2 = 1
That is, the center of the circle is (- 1.0)... You can draw the circle again
And because the guide line is x = positive and negative (a ＾ 2) / C = 0 or - 2
So, C ＾ 2 = 64, so k = 48

If the center of the circle is in the first quadrant and the circle with radius 1 is tangent to the Quasilinear of parabola y2 = 2x and the asymptote of hyperbola x216 − Y29 = 1, then the coordinates of the center of the circle are______ ．

From hyperbolic equation, we can get a = 4, B = 3, C = 5, asymptote equation y = 3x4 and y = - 3x4, that is, 3x-4y = 0 and 3x + 4Y = 0. The Quasilinear of parabola y2 = 2x is x = - 12. According to the fact that the center of the circle is in the first quadrant and the radius of 1 is tangent to the Quasilinear of parabola y2 = 2x, let the coordinates of center a be (12, m), (M > 0) When the circle is tangent to the asymptote 3x + 4Y = 0 of the hyperbola x216 − Y29 = 1, the distance from the center a to the straight line 3x + 4Y = 0 is the radius 1 of the circle, that is, | 3 × 12 + 4m | 16 + 9 = 1, {M = 78; then the coordinates of the center of the circle are: (12138) or (12, 78). So the answer is: (12138) or (12, 78)

Whether there is a point P on the line y = 3 / 2x-1, so that the circle centered on point P passes through two known points a (- 3,2), B (1,2)
Draw the picture by yourself.: -)

The distance between P and two points is equal
Let P point (3x / 2-1, y)
(3x/2+2)^2+(y-2)^2=(3x/2-2)^2+(y-2)^2
x=0 p(0,-1)