The equation of circle O is known to be x ^ 2 + y ^ 2 = 4. (1) find the tangent equation of circle O passing through point m (- 4,8); (2) pass through point n (3,0) It is known that the equation of circle O is x ^ 2 + y ^ 2 = 4. (1) find the tangent equation of circle O passing through point m (- 4,8); (2) cross point n (3,0) as a straight line with circle O at two points a and B, find the maximum area of △ OAB and the slope of line AB at this time

The equation of circle O is known to be x ^ 2 + y ^ 2 = 4. (1) find the tangent equation of circle O passing through point m (- 4,8); (2) pass through point n (3,0) It is known that the equation of circle O is x ^ 2 + y ^ 2 = 4. (1) find the tangent equation of circle O passing through point m (- 4,8); (2) cross point n (3,0) as a straight line with circle O at two points a and B, find the maximum area of △ OAB and the slope of line AB at this time

Let the tangent be Y-8 = K (x + 4), that is, kx-y + 4K + 8 = 0, the distance from the center of circle (0,0) to the straight line is radius 2, so ｜ 4K + 8 / √ (1 + K & sup2;) = 2 (4K + 8) & sup2; = 4 + 4K & sup2; 16K & sup2; + 64K + 64 = 4 + 4K & sup2; 12K & sup2; + 64K + 60 = 03k & sup2; + 16K + 15 = 0k = (- 16 ± 2 √ 19) / 6 = (- 8 ± √ 19)

It is known that the quadratic equation of one variable x2-2kx + 12k2-2 = 0 with respect to X

It is proved that: ∵ a = 1, B = - 2K, C = 12k2-2, ∵ a = 4k2-4 × 1 × (12k2-2) = 2k2 + 8, ∵ no matter what the real number k is, K2 ≥ 0, ∵ 2k2 + 8 > 0, that is ∵ 0. Therefore, no matter what the real number k is, the equation always has two unequal real roots

AC and BD are the diagonals of quadrilateral ABCD, e and F are the midpoint of AC and BD respectively, ∠ DAB = ∠ BCD = 90 °. To prove EF ⊥ AC

Connecting AF, CF
∵∠ DAB = ∠ BCD = 90 ° f is the midpoint of BD
∴AF=1/2BD,CF=1/2BD
∴FA =FC
∵ e is the midpoint of AC
∴EF⊥AC

Given that the function y = MX 2 + X + 1 has a solution when - 1 ≤ x ≤ 0, the value range of M is obtained

When m = 0, the solution of function y = x + 1, y = 0 is x = - 1, which is consistent with the problem; when m is not 0, function y = f (x) = MX ^ 2 + X + 1 is a quadratic function, y = 0, that is, f (x) = MX ^ 2 + X + 1 = 0 has a solution in [- 1,0], let the range of M be set a; now let MX ^ 2 + X + 1 = 0 have no solution in [- 1,0], and the range of M is set B, then a and B are real

In the known plane rectangular coordinate system, point O is the origin, a (- 3, - 4), B (5, - 12)
In known plane rectangular coordinate system, point O is the origin, a (- 3, - 4), B (5, - 12)
Known vector OC = (2, - 16) od = (- 8,8)
Find the cosine value of the angle between vector AC and vector BD

A (- 3, - 4) B (5, - 12) C (2, - 16) d (- 8,8) AC = (5, - 12) BD = (- 13,20) ac * BD = - 65-240 = - 305 | AC | = √ [5 & # 178; + (- 12) & # 178;] = 13 | BD | = √ [(- 13) & # 178; + 20 & # 178;] = √ 569cos = AC * BD / (- AC | BD |) = - 305 / (13 * √ 569) I'm glad to answer for you

Are there constants P and Q such that X4 + PX2 + Q can be divisible by x2 + 2x + 5? If it exists, find out the value of P and Q, otherwise, explain the reason

If there exists, then it is shown that X4 + PX2 + Q can be divided by x 2 + 2x + 5, and another factor is x2 + MX + N, and we can set another factor is x 2 + MX + N, {(x2 + 2x + 5) (x2 + 2x + 5) (x2 + 2x + 2x + 5) (x2 + MX2 + Q, that is, there is X4 + (M + 2) X3 + (M + 2 + 2m + 5) x 2 + (n + 2m + 2 + 2m + 5 + 2 + 2m + 5m) x + 5 N = n + 2m + 2m + 2 = 0n + 2m + 2m + 2m + 5 = P and 2n + 2m + 5m = 2n + 5 = 2n + 5 = 2n + 5m = 2n + 5 N = 05n = q, and 2n + 5m = 2n + 5m = 2n + 5m = 05n = 2n + 5m = 05n = 05n = 05n = 05n = q solve the above equations, and the above equations, we solve the it's not easy Find P = 6, q = 25

It is known that the edge length of the cube abcd-a1b1c1d1 is 2, e and F are the midpoint of BB1 and dd1 respectively
(2) plane ade ‖ plane b1c1f

The results show that: (1) take the midpoint of Aa1 as G, connect GF and GB1. According to the condition, GF ‖ b1c1, tangent GF = b1c1, so the quadrilateral gfc1b1 is parallelogram, then FC1 ‖ GB1, Ag ‖ EB1, and Ag = EB1, so agb1e is parallelogram, then AE ‖ GB1, so FC1 ‖ AE, get FC1 ‖ plane ade (2) FC1 ‖ AE, B

Idioms with two numbers
Clam, two numbers is OK, but it must be an idiom~

Seven up and eight down, three words, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine down, nine
Kill two birds with one stone

In a circle with a diameter of 3cm, the degree of the circumference angle of the 2 / 3cm chord is?
Sorry, I wrote the wrong question. It's a 3 / 2cm string

30 degrees and 150 degrees

How to calculate 15 × 3.5 + 3.5x-15 × 3.5-3.5x = 35

0=35
The equation doesn't hold