Given that the point a (a, radical 2) is the intersection of two functions y = kx-2 and y = (radical 2-1) x, find the real number k =? Reason

Given that the point a (a, radical 2) is the intersection of two functions y = kx-2 and y = (radical 2-1) x, find the real number k =? Reason

Ak-2 = root 2
(radical 2-1) a = radical 2
A = 2 + radical 2
K = (radical 2 + 2) / a = 1
So k = 1

Let ax-y + 3 = O and circle (x-1) 2 + (Y-2) 2 = 4 intersect at two points a and B, and the length of chord AB is 2 root sign 3, then a=

Get the center of the circle (1,2) and radius = 2
And because the length of the string AB is two, the root sign is three
So the distance from the center of the circle (1,2) to the straight line ax-y + 3 = o = under the root sign (2 ^ 2 - root sign 3 ^ 2) = 1
So the distance from the center of the circle (1,2) to the straight line ax-y + 3 = o = (A-2 + 3) / under the root sign (a ^ 2 + 1) = 1
The solution is a = 0

If the line ax-y + 3 = 0 and the circle (x-1) 2 + (Y-2) 2 = 4 intersect at two points a and B, and | ab | = 23, then a=______ ．

The center O (1,2) of the circle (x-1) 2 + (Y-2) 2 = 4, radius r = 2, passing through the center O (1,2) and making OC vertical AB, intersecting AB at point C, ∫ ab | = 23, ∫ AC | = | ab | 2 = 3, ∫ OA | = r = 2, ∫ OC | = 22 − (3) 2 = 1, that is, the distance from the center O (1,2) to the straight line AB: ax-y + 3 = 0, d = 1 | a − 2 + 3 | A2 + 1 = 1  a

Let the line x-ky-1 = 0 and the circle (x-1) ^ 2 + (Y-2) ^ 2 = 4 intersect at points a and B, and the length of chord AB is twice the root sign 3?

Center C (1,2), r = 2
From Pythagorean theorem
Distance from circle center to straight line d = √ [R & # 178; - (AB / 2) &# 178;]
therefore
|1-2k-1|/√(1+k²)=√(4-3)=1
Square on both sides
4k²=1+k²
k=±√3/3

Let the line x-my-1 = 0 intersect the circle (x-1) ^ 2 + (Y-2) ^ 2 = 4 at two points AB, and the length of the chord AB is twice the root sign 3. Then the value of the real number m is
How do you get the "center of circle (1,2)" in the first step of your solution to the above question? What formula does the following "d = | 1-2m-1 | / √ (1 + M & # 178;) = | 2m | / √ (1 + M & # 178;)" have?
I'm speechless

The standard equation of a circle with center (a, b) and radius R is (x-a) &# 178; + (y-b) &# 178; = R & # 178;; the center of circle (x-1) &# 178; + (Y-2) &# 178; = 4 is (1,2), and the distance formula from radius 2 point (m, n) to straight line ax + by + C = 0 is d = | am + BN + C | / √ (A & # 178; + B & # 178;) please refer to the fact

Let the line L: a (x + 1) + y + 2 = 0 and the circle C: (x-1) 2 + (Y-2) 2 = 4 intersect at two points a and B. (1) if the length of the chord AB is 2 root sign 3, find the value of A
(2) The value result of real number a when the area of triangle ACB is maximum

Center (1,2) radius = 2
Chord length = 2 √ 3
Then the distance from the center of the circle to the straight line = 1
|2a+2+2|/√(a^2+1)=1
A = (√ 19-8) / 3 or a = (- √ 19-8) / 3
When the area of triangle ACB is the largest, AC is perpendicular to BC, then the distance from the center of circle to the straight line = √ 2
|2a+4|/√(a^2+1)= √2
a^2+8a+7=0
(a+1)(a+7)=0
A = - 1 or a = - 7

Given that the intersection of the circle (x-3) 2 + y 2 = 4 and the line y = MX passing through the origin is p and Q, then the product of OP and OQ is [] A.5 / (1 + M & sup 2;) B.1
Given that the intersection of circle (x-3) 2 + y2 = 4 and the line y = MX passing through the origin is p and Q, then the product of OP and OQ is []
A .5／（1+m²） B .1+m² C．10 D．5

D OP * OQ = 1 * 5 the product of two straight lines passing through a point outside the circle and two points intersecting the circle is equal, right? Do you have this formula? I don't know if it's right? I've forgotten it for a long time

Given that the intersection point of circle (x-3) + y = 4 and straight line y = KX passing through the origin is p, Q, then the absolute value of OP × OQ is

Obviously, the straight line y = KX is the secant of a circle. According to the tangent secant theorem, | op | ×| OQ | is equal to the square of the tangent length of the circle passing through point O. let the center of the circle (x-3) + y = 4 be C, then there is C (3, - 1), radius r = 2, and the straight line OM and circle C are tangent to m, then cm = r = 2. In OCM, OC = (3-0) + (- 1-0) = 10, then the square of the tangent length is om = oc-cm = 10-4 = 6, that is, | op | ×| OQ | = 6

Given that the circle (x-3) ^ 2 + (y + 4) ^ 2 = 4 and the line y = KX intersect at points P and Q, then the value of OP * OQ is?
It's 21. I'm wrong.
How do you get the following formula?

Let the center of the circle be K and the radius be r, then OP * OQ = OK ^ 2-r ^ 2 = 25-4 = 21
Let OK intersect at two points m n
According to the cutting line theorem, we know OP * OQ = om * on. And OM = ok-r, on = OK + R

Given that the circle (x-3) ^ 2 + (y + 4) ^ 2 = 4 and the line y = KX intersect P and Q, then the value of | op | * | OQ |?

Only for ideas!
Have you learned the cutting line equation of circle?
OP | * | OQ | is the product of two parts of a secant, equal to the square of the tangent
Because the distance from the center of the circle to o is 5 and the radius is 2, the tangent length is 23
So the answer is 23