Find the equation of the circle passing through point a (0,6) and tangent to the origin of circle C: x ^ 2 + y ^ 2 + 10x + 10Y = 0 Can you be more specific?

Circle C: x ^ 2 + y ^ 2 + 10x + 10Y = 0
(x+5)^2+(y+5)^2=50
C(-5,-5),r=5√2
Let the equation of the circle l be:
(x-a)^2+(y-b)^2=R^2
Circle l passes through point a (0,6)
a^2+(6-b)^2=R^2.(1)
Circle L and circle C: x ^ 2 + y ^ 2 + 10x + 10Y = 0 tangent to the origin
a^2+b^2=R^2.(2)
LC=R+r=R+5√2
(a+5)^2+(b+5)^2=LC^2=(R+5√2)^2.(3)
By solving the equations (1), (2), (3), we get
a=3,b=3,R=3√2
The equation of the circle L is: (x-3) ^ 2 + (Y-3) ^ 2 = 18

How to solve x + Y > 11,10x + y-18 = 10Y + 10x

10x y-18-10y-10x=0——9y-18=0——9y=18——y=2 x y＞11——x 2＞11——x＞9

It is known that the two roots of the quadratic equation (mx-n) of one variable with respect to X are X1 = 1, X2 = 2 (1) m, N, and the relation (2) M = 1. The above equation is written as a general form of quadratic equation of one variable

1. Substitute x = 1, x = 2 into
So (m-n) ^ 2 = 2
(2m-n)^2=2
∵m≠0
∴2m-n=-m+n
So 3M = 2n
2. Substitute M = 1 into n = 2 / 3
We get (X-2 / 3) ^ 2-2 = 0
x^2-3x-1/2=0

The solution equation: X △ 1 / 5 = 6.4 X-9 / 4x = 4.5 △ 2 / 7

X △ 1 / 5 = 6.4;
x÷(6/5)=6.4;
x=(6/5)×6.4;
x=7.68;
4X of X-9 = 4.5 △ 2 and 1 of 7
5x/9=4.5÷(15/7);
5x/9=2.1;
x=3.78;

Simplification: how much is x-178; + 4x + 4 / x-178; - 4 minus X / X-2?

The original formula = [(X & # 178; + 4x + 4-x (x + 2)] / (x + 2) (X-2)
=(x²+4x+4-x²-2x)/(x+2)(x-2)
=2(x+2)/(x+2)(x-2)
=2/(x-2)

Is 1 / 3x + 1 = 2x a linear equation with one variable

No, it's a fractional equation

A differentiable function f (x) becomes f * (x) after deriving. F * (x) is still a function of X. f * (x) may not be continuous! Then f * (x) may not be continuous everywhere?
Whether there is a differentiable function whose derivative is discontinuous everywhere

Continuous function sequence can only converge to a function whose discontinuous point set is the first class set, while real number set is the second class set
So it doesn't exist

Factorization: am ^ 2-4a and a ^ 2 + 12a + 36

am^2-4a
=a(m^2-4)
=a(m+2)(m-2)
a^2+12a+36
=(a+6)^2

Solve the inequality | 2x + 1 | + | X-2 | > 4

When x ≤ - 12, the original inequality can be reduced to - 2x-1 + 2-x ＞ 4, X ＜ - 1. When - 12 ＜ x ≤ 2, the original inequality can be reduced to 2x + 1 + 2-x ＞ 4, X ＞ 1. When x ＞ 2, the original inequality can be reduced to 2X + 1 + X-2 ＞ 4, X ＞ 53, X ＞ 2, X ＞ 2. In conclusion, the solution set of the original inequality is {x | x ＜ - 1 or 1 ＜ x}

If X1 and X2 are two zeros of the function y = x - (K-2) x + (K + 3K + 5), (K ∈ R), then the maximum value of X1 + X2 is

X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2, because X1 and X2 are two zeros of function y, so X1 + x2 = - B / A, x1x2 = C / A, so X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2 = (K-2) ^ 2-2 (k ^ 2 + 3K + 5) = - K ^ 2-10k-6, so the maximum value of X1 + X2 is - K ^ 2-10k-6, which is - 31

Find the equation of the circle passing through point a (0,6) and tangent to the origin of circle C: x ^ 2 + y ^ 2 + 10x + 10Y = 0 Can you be more specific?