High school mathematics -- linear equation of circle It is known that the circle C: x ^ 2 + (Y-1) ^ 2 = 5 and the straight line L: mx-y + 1-m = 0 Q: let the line L intersect the circle C at two points a and B. If | ab | = 17 under the root sign, calculate the inclination angle of L The reference answer is π / 3 or (2 π) / 3 How to calculate, I can know the linear l constant (1,1)

High school mathematics -- linear equation of circle It is known that the circle C: x ^ 2 + (Y-1) ^ 2 = 5 and the straight line L: mx-y + 1-m = 0 Q: let the line L intersect the circle C at two points a and B. If | ab | = 17 under the root sign, calculate the inclination angle of L The reference answer is π / 3 or (2 π) / 3 How to calculate, I can know the linear l constant (1,1)

Center of circle (0,1), radius √ 5, make CD perpendicular to ab through center of circle C, then CD is the distance from C to ab = | 0-1 + 1-m | / √ (m ^ 2 + 1) = | m | / √ (m ^ 2 + 1) d is the midpoint of AB, ad = √ 17 / 2ca = r = √ 5, so by Pythagorean theorem Ca ^ 2 = CD ^ 2 + ad ^ 25 = 17 / 4 + m ^ 2 / (m ^ 2 + 1) m ^ 2 = 3, slope is m, so tan angle = √ 3 or - √ 3, so

High school mathematics -- the combination of linear equation and circular equation
Given that the line L passes through point P (1,1) and intersects with the line M: X-Y + 3 = 0 and N: 2x + y-6 = 0 at points a and B respectively, if the line AB is bisected by point P, find:
1. The equation of line L
2. The equation of the circle whose chord length is (8 √ 5) / 5 is obtained by taking the coordinate origin o as the center and being cut by L

Now that we know the coordinates of point P, we can make a formula, and let the slope be K, we can get the equation y = K (x-1) + 1, so we can get the coordinates of two points a and B. because P is the midpoint, we can get the slope k = - 1 / 2, so the equation is x + 2y-3 = 02

Is the equation of line or circle difficult in high school mathematics?

These two are not too difficult, it seems that the hyperbola behind is more difficult
But to compare these two, I think it's a straight line difficulty, because it always has the accumulation of symmetrical incident reflection, and it's a very troublesome problem. The circle is mainly combined with graphics