In polar coordinate system, given that the center of circle C is C (1, π 4) and the radius is 1, the polar coordinate equation of circle C is___ .

In polar coordinate system, given that the center of circle C is C (1, π 4) and the radius is 1, the polar coordinate equation of circle C is___ .

The center of circle C is C (1, π 4), i.e. (22, 22), radius is 1. The equation of circle C is (x-22) 2 + (y-22) 2 = 1. It is changed into x2-2x + y2-2y = 0, it is changed into ρ 2-2 ρ cos θ - 2 ρ sin θ = 0, i.e. ρ = 2 (sin θ + cos θ) = 2cos (θ - π 4). So the answer is: ρ = 2cos (θ - π 4)
Polar coordinate equation of a circle with radius 2 and center at the pole
ρ=2
Polar coordinate equation of circle with center at pole O and radius a
ρ = 3cos θ + 3 √ 3sin θ can also be written as ρ = 6sin (θ + π / 6)
The polar coordinate pole (1.1) is the center of the circle, 1 is the equation of the radius of the circle
cosine law
It can be obtained that (1.1) is the center of the circle and 1 is the radius of the circle, and the equation is p = 2cos (A-1)
Let a be a real constant, and f (x) = LG (2 / (1-x) + a) be an odd function, and solve the inequality f (x)
If f (x) = 0, then f (x) = 0
f(0)=lg(2/(1-0)+a)=lg(2+a)=0
SO 2 + a = 1, a = - 1
f(x)=lg[2/(1-x)-1]=lg[(1+x)/(1-x)]
f(x)
A is a real constant and f (x) = LG (2 / (1-x) + a) is an odd function
Then f (0) = 0
lg(2/(1-0)+a)=0
a=-1
f(x)
Given the sum of the first n terms of the sequence an and Sn = 3 ^ n - 1, the sequence BN satisfies B1 = 1, BN = 3B (n-1) + an, note that the sum of the first n terms of the sequence BN is TN. find TN
Solution;
When n = 1, A1 = S1 = 3-1 = 2
When n ≥ 2, an = SN-S (n-1) = 3 & # 8319; - 1-3 ^ (n-1) + 1 = 2 × 3 ^ (n-1)
When n = 1, A1 = 2 × 1 = 2, which also satisfies the general formula
The general term formula of sequence {an} is an = 2 × 3 ^ (n-1)
bn=3b(n-1)+2×3^(n-1)
Divide both sides of the equation by 3 & # 8319;
bn/3ⁿ=b(n-1)/3^(n-1) +2/3
BN / 3 &; - B (n-1) / 3 ^ (n-1) = 2 / 3, which is the fixed value
B1 / 3 = 1 / 3, the sequence {BN / 3 & # 8319;} is an equal ratio sequence with 1 / 3 as the first term and 2 / 3 as the common ratio
bn/3ⁿ=(1/3)(2/3)^(n-1)=2^(n-1)/3ⁿ
bn=2^(n-1)
Tn=b1+b2+...+bn=1+2+...+2^(n-1)=1×(2ⁿ-1)/(2-1)=2ⁿ-1
As shown in the figure, in △ ABC, the linear equation of the height on the BC side is x-2y + 1 = 0, and the linear equation of the bisector of ∠ A is y = 0. If the coordinates of point B are (1,2), the coordinates of point a and point C are obtained
Point a is the intersection of two lines y = 0 and x-2y + 1 = 0, the coordinates of point a are (- 1, 0).. KAB = 2 − 01 − (− 1) = 1. The equation of the bisector line of ∵ - A is y = 0, ∵ - KAC = - 1. The equation of line AC is y = - X-1. BC is perpendicular to x-2y + 1 = 0, ∵ KBC = - 2. The equation of line BC is Y-2 = - 2 (x-1). From y = - X-1, y = - 2x + 4, the solution of C (5, - 6).. the coordinates of point a and point C are obtained They are (- 1,0) and (5, - 6) respectively
X & # 178; - 2x - (m-2) = 0 and X & # 178; + MX + 1 / 4m & # 178; + m + 2 = 0, if the two equations have at least one real solution, the range of M is obtained
thanks.
When the triangle is equal to zero, there is one. If the triangle is greater than zero, there is at least one real solution of the equation, that is, finding the intersection of two equations can get the range of M!
Let f (x) = LG (2 / 1-x + a) be an odd function and solve the inequality f (x)
Odd function
f(0)=0
So a = - 1
f(x)=lg[(1+x)/(1-x)]
Because the function is an odd function, we have: F (0) = 0, and substitute it to get: LG (2 + a) = 0. So a = (x + 1) / (1-x) 1 or X
Given that the sequence {an} satisfies Sn = N2 + 1, the sequence {BN} satisfies BN =, and the sum of the first n terms is TN, let CN = T2N + 1-tn
(1) ∵ sequence {an} satisfies the first N-term sum Sn = n square + 1 ∵ Sn = n ^ 2 + 1 s (n-1) = (n-1) ^ 2 + 1 an = SN-S (n-1) = n ^ 2 + 1 - [(n-1) ^ 2 + 1] = 2N-1 A1 = S1 = 2 BN = 2 / an + 1 = 2 / (2n-1) + 1 = (2n + 1) / (2n-1) B1 = 2 / A1 + 1 = 2 BN is a sequence with the first term of 2 and the general term of (2n + 1) / (2n-1)