Find the equation of the circle whose center is on the line 3x-y = 0, tangent to the x-axis, and cut by the line X-Y = 0

Find the equation of the circle whose center is on the line 3x-y = 0, tangent to the x-axis, and cut by the line X-Y = 0

If the center of the circle is on the line 3x-y = 0 = = = > y = 3x, let the coordinates of the center of the circle be (m, 3M),
If it is tangent to the x-axis, then the radius is 3 m, and the circle equation is: (x-m) 2 + (Y-3 m) 2 = 9 m
If the chord center distance = D, then according to Pythagorean theorem, D 2 = (3 m) 2 - [(2 √ 7 / 2)] 2 = 9 m 2 - 7,
∵ according to the distance formula from the point (m, 3M) to the straight line 3x-y = 0, d = | m-3m | / √ 2 = √ 2m,
∴(√2m)²=9m²-7===>7m²=7,===>m=±1,
Then the circle equation is:
(x-1) 2 + (Y-3) 2 = 9, or: (x + 1) 2 + (y + 3) 2 = 9

The chord length of the line L: y = X-1 cut by circle C is 2 2, then the equation of the line passing through the center of the circle and perpendicular to the line L is () A. x+y+1=0 B. x+y-1=0 C. x+y-2=0 D. x+y-3=0

Let the coordinates of the center of the circle be (a, 0), then
The chord length of the line L: y = X-1 cut by the circle is 2
2 is (| a − 1)|
2) 2 + 2 = (A-1) 2, a = 3 or - 1,
Because the center of the circle is on the negative half axis of the X axis, a = - 1, so the coordinates of the center of the circle are (- 1, 0),
∵ the slope of the line L is 1
The equation of the line passing through the center of a circle and perpendicular to the line L is y-0 = - (x + 1), that is, x + y + 1 = 0
Therefore, a

Given that the circle C passes through the point (1, 0), and the center of the circle is on the positive half axis of the X axis, the chord length of the straight line L: y = X-1 cut by the circle is 2 2, then the standard equation of circle C is () A. （x+1）2+y2=4 B. （x-3）2+y2=4 C. （x-1）2+y2=4 D. （x+3）2+y2=4

Let the coordinates of the center of the circle C be (a, 0), a > 0, then the distance from the center of the circle to the straight line L: y = X-1 is d = | a − 0 − 1|
2=|a−1|
2．
Because the radius r = | A-1|=
(a−1
2)2+(
2) 2, a = 3, or a = - 1 (omitted),
Therefore, the center of circle C is (3,0), and the radius is 3-1 = 2. Therefore, the standard equation of circle C is (x-3) 2 + y2 = 4,
Therefore, B

It is known that a circle and a straight line 3x + 4Y-2 = 0 are tangent to point P (2, - 1), and the length of chord obtained by cutting the positive half axis of X axis is 8. The standard equation of this circle is obtained

Let the circle equation be (x-a) 2 + (y-b) 2 = R2 passing through the point P, and the line perpendicular to the line 3x + 4Y-2 = 0 is y + 1 = 43 (X-2), that is, 4x-3y-11 = 0, and the distance from the center of the circle (a, B) to the point P is d = R, that is, 4a-3b-11 = 0, ① | 3A + 4B − 2 | 32 + 42 = R. ② substituting y = 0 into the circle equation, we can get: (x-a)

It is known that the center of a circle C is (2, - 1), and the chord length of the circle cut by the straight line L: x-y-1 is twice the root sign 2, and the equation of the circle is solved

Let the radius of the circle be r, make a vertical line from the center of the circle to the straight line L, and intersect the point d (1,0)
Let l intersect circle C at points a and B, then connect Ca and CB
Then the height of AB edge of △ ABC is CD = √ 2, CA = CB = R, ab = 2 √ 2
So r = CD ^ 2 + (AB / 2) ^ 2 = 2
The radius is 2
The circle equation is
(x-2)^2+(y+1)^2=4

It is known that: the center of a circle is C (2, - 1), and the chord length of the circle cut by the line L: x-y-1 = 0 is twice the root sign 2. Find the equation of the circle

The distance from the center of the circle to the straight line is | 2 - (- 1) - 1 | / √ (1 + 1) = 2 / √ 2 = √ 2
Half mystery, radius, the distance from the center of a circle to a straight line
If the radius of a circle is r, then R ^ 2 = (2 √ 2 / 2) ^ 2 + (√ 2) ^ 2 = 2 + 2 = 4
So the equation of the circle is (X-2) ^ 2 + (y + 1) ^ 2 = 4

A circle is tangent to the y-axis, and the chord length cut on the line x-3y = 0 on y = x is 2 times the root sign 7, and the circle equation is solved

Let the coordinates of the center of a circle (a, b) because the circle C is tangent to the Y axis, so r = A. according to the geometric relationship, d ^ 2 = R ^ 2 - (√ 7) ^ 2 can be obtained. According to the formula of distance from point to line, D ^ 2 = [(a-b) ^ 2] / 2, so R ^ 2 - (√ 7) ^ 2 = [(a-b) ^ 2] / 2

It is known that a circle and the straight line 3x + 4Y-2 = 0 are tangent to the point P (2, - 1), and the length of the chord obtained by cutting the positive half axis of the X axis is 8. The standard equation of the circle is obtained______ .

Let C (a, b) be the center of the circle C (a, b), PC and the straight line 3x + 4Y-2 = 0 are vertical, and the chord center distance, half chord length and radius form a right triangle, so there is B + 1a − 2 × (− 34) = − 1b2 + 16 = (a − 2) 2 + (B + 1) 2

Given that the center of the circle is on the line x-y-1 = 0 and tangent to the line 4x + 3Y + 14 = 0, and the line 3x + 4Y + 10 = 0, the equation of the circle with chord length of 6 is obtained

First of all, if the center of the circle is on the line L1: x-y-1 = 0, then we may as well set the coordinates of the center of the circle as (a, A-1)
Because the circle is tangent to the straight line l2:4x + 3Y + 14 = 0, then it is obtained from the distance formula of point to line
Radius r = D = [4A + 3 (A-1) + 14] / 5 = (7a + 11) / 5
If the chord length of a circle cut on the straight line l3:3x + 4Y + 10 = 0 is 6, then the formula of distance from point to line is obtained
The distance from the center of a circle to L3 is d = [3A + 4 (A-1) + 10] / 5 = (7a + 6) / 5
From this, you draw a graph, observe, make the vertical line of L3 through the center of the circle, and the chord length is divided into two
The square of half the chord length L + the square of D = the square of R
So we can get the relationship: (6 / 2) ^ 2 + [7a + 6 / 5] ^ 2 = [7a + 11 / 5] ^ 2
By solving the above equation, a = 2 is obtained
So the center of the circle is r = 1
The equation of the circle is (X-2) ^ 2 + (Y-1) ^ 2 = 25

It is known that a circle and a straight line 3x + 4Y-2 = 0 are tangent to point P (2, - 1), and the length of chord obtained by cutting the positive half axis of X axis is 8. The standard equation of this circle is obtained

Let the circle equation be (x-a) 2 + (y-b) 2 = R2
The line passing through point P and perpendicular to the line 3x + 4Y-2 = 0 is y + 1 = 4
3 (X-2), that is 4x-3y-11 = 0
The center of the circle (a, b) is on the straight line, and the distance from the center to the point P is d = R
4a-3b-11=0，①
|3a+4b−2|
32+42=r②
Substituting y = 0 into the circular equation, we get: (x-a) 2 + B2 = R2,
X 1 = a+
r2−b2，x2=a-
r2−b2
The chord length 8 = | x1-x2| = 2 is obtained by cutting the positive half axis of X axis
R2 − B2, that is, r2-b2 = 16
① (2) and (3) the simultaneous solution is: r = 5, a = 5, B = 3
So the circle equation is (X-5) 2 + (Y-3) 2 = 25