The equation of a straight line passing through point P (2,1) and tangent to circle x ^ 2 + y ^ 2-2x + 2Y + 1 = 0 is What's the first thing to think about when doing this kind of problem

The equation of a straight line passing through point P (2,1) and tangent to circle x ^ 2 + y ^ 2-2x + 2Y + 1 = 0 is What's the first thing to think about when doing this kind of problem

Formula: (x-1) ^ 2 + (y + 1) ^ 2 = 1
The center of the circle is (1, - 1) and the radius is 1
There is obviously a vertical tangent: x = 2
Let another tangent be y = K (X-2) + 1
Then the distance from the center of the circle to the straight line is equal to the radius
|k(1-2)+1+1|^2/(1+k^2)=1
That is: (K-2) ^ 2 = 1 + K ^ 2
k=3/4
So the other tangent is y = 3 (X-2) / 4 + 1 = 3x / 4-1 / 2
Let Y-1 = K (X-2)
Get the coordinates of the center of the circle
The distance from the center of the circle to the straight line is equal to the radius
It's K
It's OK
(XY-2 / 3y-2 / 1) - (XY-2 / 3x + 1), where x = 10 / 3, y = 8 / 3
Original formula = XY-2 / 3y-2 / 1-xy + 2 / 3x-1
=-3Y / 2 + 3x / 2 -- 3 / 2
=-4 + 5-3 / 2
=-1 / 2
4: What is the positional relationship between circle C1: x2 + y2-2x = 0 and circle C2: x2 + y2-4y = 0?
C1：(x-1)²+y²=1；C2：x²+(y-2)²=4
C1(1,0),C2(0,2)
r1=1,r2=2,
Center distance d = C1C2 = √ 5
r2-r1=1,r1+r2=3
r2-r1
Banana
Mix X in C1 and Y in C2 into formula.
Arrange the standard form of a circle, and draw on the paper according to the center and radius of the circle.
Find 3x + XY + 3Y of x-xy + y, where x + y of xy = 3
X + y = 3
x+y=3xy
3x + XY + 3Y of x-xy + y
=[3(x+y)+xy]/[(x+y)-xy]
=(9xy+xy)/(3xy-xy)
=10xy/2xy
=5
Because: x + y = 3
So x + y = 3xy
Substituting x + y = 3xy into 3x + XY + 3Y of x-xy + y
The result is: 3x + XY + 3Y = 10xy / 2XY = 5
Let the line 3x + 4y-5 = 0 and circle C1: x2 + y2 = 4 intersect at two points a and B. if the center of circle C2 is on line AB, and circle C2 is tangent to circle C1, and the tangent point is on the inferior arc AB of circle C1, then the maximum radius of circle C2 is______ .
From the circle C1: x2 + y2 = 4, we can get the center O (0, 0) and radius r = 2. As shown in the figure, when the center Q of circle C2 is the midpoint of line AB, circle C2 is tangent to circle C1, and the tangent point is on the inferior arc AB of circle C1. Let p be the tangent point, then the radius r of circle C2 is the largest
The following equations 8y + 5x = 2,4y-3x = - 10 are solved by the method of addition and subtraction,
8y＋5x=2①；
4y-3x=-10②；
② * 2, 8y-6x = - 20 ③;
① - 3, we get: 11x = 22, we get: x = 2;
Replace the original equation and get y = - 1
It is known that circle C1: x ^ 2 + y ^ 2 + 3x-4 = 0 and circle C2: x ^ 2 + y ^ 2 + 4Y = 0 intersect at two points ab
Then the chord length of C3: X ＾ 2 + y ＾ 2-2x-2y-7 = 0 is?
X ^ 2 + y ^ 2 + 3x-4 = 0 and x ^ 2 + y ^ 2 + 4Y = 0 are ab straight lines, so the straight line is 3x-4y-4 = 0, and then y = (3 / 4) X-1 is substituted into circle 3 to get 25X ^ 2-80x-64 = 0, because d = radical [(x1-x2) ^ 2 + (y1-y2) ^ 2] = | x1-x2 | * radical (1 + K ^ 2) = radical (1 + K ^ 2) * radical [(x1 + x2) ^ 2-4x1 * x
3x + 4Y = - 5 5x + 6y = - 9 (n-8m) x-8y = 10 5x + (10m + 2n) y = - 9 for M.M + N.N =?
117/169
It is known that AB is the diameter of ⊙ o, AC is the chord of ⊙ o, and point D is the midpoint of ABC. The chord de ⊥ AB is at point F, and de intersects AC at point g. (1) as shown in Fig. 1, it is proved that ⊙ BAC = ⊙ OED; (2) as shown in Fig. 2, the tangent of ⊙ o is made through point E, and the extension line of AC is at point h. if AF = 3, FB = 43, the value of cos ⊙ DEH is obtained
(1) It is proved that: connect do and extend AC to point m, as shown in Fig. 1, ∵ point D is the midpoint of ABC, ∵ om ⊥ AC, ∵ amo = 90 °, ∵ de ⊥ AB, ∵ ofd = 90 °, and ∵ AOM = ∵ DOF, ∵ a = ∵ D, ∵ od = OE, ∵ OED = ∵ D, ∵ BAC = ∵ OED; (2) connect OE, as shown in Fig. 2, ∵ EH is ⊙ o
Solve the system of binary equations about X, Y: 3x + 4Y = - 55x + 6y = - 9 (n-8m) x-8y = 105x + (10m + 2n) y = - 9 find the sum of squares of M + n
Through the first two, we can get the solution of X. y and bring it into the following two formulas to solve the binary linear equation of M and n. We can get the solution of M and N, x = - 3, y = 1, M = 9 / 13, n = - 6 / 13, sum of squares = 145 / 169