The equation of the circle whose center is on the line 2x + y = 0 and tangent to the point (2, - 1) with the line x + Y-1 = 0 is______ ．

The equation of the circle whose center is on the line 2x + y = 0 and tangent to the point (2, - 1) with the line x + Y-1 = 0 is______ ．

Let the center coordinates of the circle be (a, b), then 2A + B = 0|a + B − 1|2 = & nbsp; (a − 2) 2 + (B + 1) 2, the solution is a = 1, B = - 2, so r = 2, so the equation of the circle is (x-1) 2 + (y + 2) 2 = 2

Given that the circle passes through point a (2, - 1), the center of the circle is on the line 2x + y = 0 and tangent to the line x-y-1 = 0, the equation of the circle is obtained

Let the equation of the circle be (x-a) 2 + (y-b) 2 = R2 (r ). (the center of a circle is on the line 2x + y = 0, and \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\132 The equation of circle is (x-1) 2 + (y + 2) 2 = 2 or (X-9) 2 + (y + 18) 2 = 338

The circle passes through P (2, - 1), is tangent to the straight line X-Y = 1, and its center is on y = - 2x

The center of the circle is on y = - 2x
Let the radius of the center O (A. - 2A) be r, then the equation of the circle is (x-a) ^ 2 + (y + 2a) ^ 2 = R ^ 2
The distance from point O to the straight line X-Y = 1 is r, i.e
(3a-1)^2 /2 = r^2.(1)
And Po = R, with
(a-2)^2 + (-2a+1)^2 = r^2.(2)
From (1), (2)
(3a-1)^2 /2 = (a-2)^2 + (-2a+1)^2
Reduced to a ^ 2 - 10A + 9 = 0
The solution is a = 1 or 9
When a = 1, the equation of R ^ 2 = 2 circle is (x-1) ^ 2 + (y + 2) ^ 2 = 2
When a = 9, the equation of R ^ 2 = 338 is (X-9) ^ 2 + (y + 18) ^ 2 = 338

Find the equation of the circle with the intersection of two straight lines L1: X-Y = 5 and L2: 2x + y = 4 as the center and tangent to the X axis

x-y=5
2x+y=4
Solution x = 3, y = - 2
So the coordinates of the center of the circle are (3 - 2)
Because the circle is tangent to the X axis
So the radius is two
The equation of a circle is (x-3) & sup2; + (y + 2) & sup2; = 4

It is known that three straight lines L1: x-2y = 0, L2: y + 1 = 0, L3: 2x + Y-1 = 0 intersect each other, and the equation of the circle of these three intersections is solved______ ．

The coordinates of the three intersections are (- 2, - 1), (1, - 1), (25, 15) respectively. Let the equation of the circle be x2 + Y2 + DX + ey + F = 0. According to the circle passing through the three intersections, we can get 4 + 1 − 2D − e + F = 01 + 1 + D − e + F = 0425 + 125 + 2D5 + E5 + F

As shown in the figure, AB is the diameter of circle O, C is the point on circle O, D is the midpoint of arc AC, de ⊥ AB is perpendicular to e, and AC intersects de and dB at points F and g respectively. Are AF and FG equal? Why?

AF = FG, the reason is: connecting ad, ∵ AB is diameter, de ⊥ AB, ∵ ADB = ∵ DEB = 90 °, ∵ ade = ∵ abd, ∵ D is arc AC midpoint, ∵ DAC = ∵ abd, ∵ ade = ∵ DAC, ∵ AF = DF, ∵ FAE = ∵ DAC, ∵ DF = FG, ∵ AF = FG

In ⊙ o, the chord AB = AC = 10, the chord ad intersects BC with E, AE = 4, and the length of ad is obtained
Chapter 7 of basic training p23
(if the title is wrong, please indicate how to do it)

Connect BD
∵AB=AC
Ψ arc AB = arc AC
∴∠D =∠ABC
∵∠BAE=∠DAB
∴△ABE∽△ADB
∴AB/AD =AE/AB
∴AB²=AE*AD
That is, 100 = 4AD
∴AD=25

If a is a point on a circle, a and O intersect B and C, and ab = 4, and the chord ad of O intersects BC with E, then what is ad multiplied by AE

Connecting BD and AC
Because AB = AC, so angle ABC = angle ACB; and because angle ACB = angle D, so angle ABC = angle D, so △ AEB ∽ abd, then we can get ad: ab = AB: AE, that is, ad × AE = ab × AB = 16

Take a point E on the extension line of the chord CD perpendicular to the diameter AB, connect the intersection circle of AE to F, and verify: AC * EF = de * CF

Because the vertical diameter of string CD is ab
So arc ad = arc AC
So angle DFA = angle AFC
Because arc FC = arc FC
So angle fac = angle FDC
Because angle DFA = angle AFC, angle fac = angle FDC
So triangle ACF is similar to triangle def
So AC / CF = de / ef
So ac * EF = de * CF

AB, CD are the diameter of ⊙ o, AE, CD are chords, and AE = CF, the proof is: ⊙ a = ⊙ C

Connecting EB, CF
⊙ AB, CD is the diameter of ⊙ o
∴∠AEB=∠CFD
∵AE=CF
∴∠ABE=∠CDF
∴ ∠A=180- ∠AEB -∠ABE
∠C=180- ∠CFD -∠CDF
∴∠A=∠C