A beam of light is emitted from a (- 2,3), reflected by x-axis, and tangent to the circle C (x-3) 2 + (Y-2) 2 = 1. The equation of the straight line of the reflected light is obtained Method 1: the symmetrical point of point a about the x-axis is on the line where the reflected light is located PS: it's a clue to the title, but I'm not good at it

A beam of light is emitted from a (- 2,3), reflected by x-axis, and tangent to the circle C (x-3) 2 + (Y-2) 2 = 1. The equation of the straight line of the reflected light is obtained Method 1: the symmetrical point of point a about the x-axis is on the line where the reflected light is located PS: it's a clue to the title, but I'm not good at it

Let's set up the equation of undetermined coefficients of reflection, and use the tangent condition to get an equation of coefficients. Let's make 1. For X reflection, we change y into - y, and then substitute the known points into the equation of coefficients. 2. Let's set 1 and 2 together to get the coefficients, then we can get the equation of reflection

The equation of a line with chord length 8 passing through point P (3,6) and cut by circle x2 + y2 = 25 is______ ．

If the slope of the straight line does not exist, then the vertical X-axis x = 3, the distance from the center of the circle to the straight line = | 0-3 | = 3. If the slope exists y-6 = K (x-3), that is: kx-y-3k + 6 = 0, then the distance from the center of the circle to the straight line | 0 − 0 − 3K + 6 | K2 + 1 = 3. The solution is k = 34. In summary: x-3 = 0 and 3x-4y + 15 = 0, so the answer is: x-3 = 0 and 3x-4y + 15 = 0

Given that a straight line passes through the point P (- 3, - 3 / 2) and intersects the circle x ^ 2 + y ^ 2 = 25, the chord length obtained is 8, and the linear equation?

The distance from the center of the circle to the straight line = √ (5 ^ 2-4 ^ 2) = 3 if the chord length of a straight line intersecting the circle is 8 and its radius is 5, passing through the point P (- 3, - 3 / 2). Therefore, when the slope of the straight line does not exist, the straight line equation x = - 3 if the slope of the straight line exists, let the straight line equation be y + 3 / 2 = K

Find the linear equation of six 2-long strings parallel to 3x + 3Y + 5 = 0 and cut by circle x ^ 2 + y ^ 2 = 20
Please be more detailed

Let the straight line be y = KX + B
∵ the line is parallel to 3x + 3Y + 5 = 0
The slope of the straight line k = - 1
That is, the straight line is y = - x + B
And the straight line intersects the circle,
∴x^2+(-x+b)=20
2x^2-2bx+b^2-20=0
Then x2-x1 = b
And x = - y + B
∴2y^2-2by-20=0
Then y2-y1 = b
The chord length of the circle cut by the straight line is 6 √ 2;
∴√[(x2-x1)^2+(y2-y1)^2]=6√2
Namely: √ 2B ^ 2 = 6 √ 2
b^2=6
b=±6
The straight line is y = - X-6 or y = - x + 6