The curve represented by equation x (x ^ 2 + y ^ 2-4) = 0 and x ^ 2 + (x ^ 2 + y ^ 2-4) ^ 2 = 0 A represents a straight line and a circle, B represents two points C the former is a line and a circle, the latter is two points, d the former is two points, the latter is a line and a circle

The curve represented by equation x (x ^ 2 + y ^ 2-4) = 0 and x ^ 2 + (x ^ 2 + y ^ 2-4) ^ 2 = 0 A represents a straight line and a circle, B represents two points C the former is a line and a circle, the latter is two points, d the former is two points, the latter is a line and a circle

x(x^2+y^2-4)=0
We know that x = 0 or x ^ 2 + y ^ 2-4 = 0
Represents a line and a circle
x^2+(x^2+y^2-4)^2=0
There must be
x^2=0
(x^2+y^2-4)^2=0
The solution is x = 0, y = ± 2
Represents two points
So choose C

It is known that the two intersection points of square of circle x + square of Y + x-6y + C = 0 and straight line x + 26y-3 = 0 are p and Q, and op ⊥ OQ (o is the origin) is the equation for solving circle

First draw a picture, change the equation of the circle into a standard equation, and calculate: the coordinates of the center of the circle are (- 0.5,3), and let the center of the circle be m
The distance between M and PQ (PQ equation, i.e. line x + 26y-3 = 0) can be obtained by the formula of distance from point to line = (you calculate), and I will replace it with the letter D
The equations are composed of the equations of circle and line
X+26Y-3=0
Square of X + square of Y + x-6y + C = 0
Finally, two sets of solutions can be obtained: x = ,Y=… The solutions of. X and Y contain the unknown number C
The solutions of X and y are the coordinates of P and Q respectively
Then the distance formula between two points is used to get: PQ = The root sign can be eliminated after the square, which is convenient for the next step
Because the distance between M and PQ is perpendicular to PQ, connect MP, then MP is radius R (you can see clearly by drawing)
With Pythagorean theorem, the square of R (after the equation of circle is transformed into standard equation, we can also know that R contains unknown number C) = the square of half of PQ + the square of D
Simplify the above formula to form an equation with only unknown number C, and then solve C
Finally, we substitute C into the square of the circle x + the square of Y + x-6y + C = 0
complete

The equation of circle
The equation of circle is X & # 178; + Y & # 178; = 1, and the equation of straight line mx-y + 2 = 0. When m is the value, the circle and the straight line are tangent, intersect and separate? This is a preview study plan. I really don't understand it

The equation of circle is X & # 178; + Y & # 178; = 1, and the equation of straight line mx-y + 2 = 0. When m is the value, the circle and the straight line are tangent, intersect and separate?
If the center of the circle X & # 178; + Y & # 178; = 1 is at the origin (0,0) and the radius r = 1, and the distance from the center of the circle to the straight line mx-y + 2 = 0 is D, then:
D = 2 / √ (M & # 178; + 1), when d = 2 / √ (M & # 178; + 1) = 1, that is, M & # 178; + 1 = 4, M & # 178; = 3, M = ± √ 3, the line is tangent to the circle;
When d = 2 / √ (M & # 178; + 1) 4, M & # 178; > 3, M > √ 3 or M1, that is M & # 178; + 1 > 4, M & # 178;