Given the straight line L1: 3x-4y-17 = 0, L2: 3x-4y + 23 = 0, find the equation of circle C tangent to L1, L2 and X axes

Given the straight line L1: 3x-4y-17 = 0, L2: 3x-4y + 23 = 0, find the equation of circle C tangent to L1, L2 and X axes


The distance between L1 and L2 is d = (23 + 17) / 5 = 8, and the circle is sandwiched between the parallel lines L1 and L2. Therefore, the diameter is 8.. r = 4 and the center of the circle is on the line L3 in the middle of L1 and L2: 3x-4y + 3 = 0. R = 4 and tangent to the X axis. Then the ordinate of the center of the circle is positive or negative 4



1. The equation of circle tangent to the line 3x-4y-17 = 0, 3x-4y + 23 = 0 and x-axis
I've basically listed three equations, but I can't seem to work out the result


In two parallel lines, the center of the circle is on a straight line. Find out the line 3x + 4Y + 3 = 0, and then find out the intersection point with X ˊ. The point of this point 4 on this straight line is the center radius of the circle, which is 4. You can calculate it slowly



If a straight line L passes through point a (- 1, - 3) and its inclination angle is equal to twice the inclination angle of the straight line y = 2x, then the linear equation is equal to?


k1=tana1=2
k2=tan(2a1)=2tana1/(1-(tana1)^2)=4/(1-4)=-4/3
y=-4/3*(x+1)-3
4x+3y+13=0



It is known that the line m and the line L: 2x-y + 3 = 0 have the same intercept on the Y axis, and the inclination angle of the line m is twice that of the line L. The equation of the line m is obtained


If the slope of line L is - A / b = 2 and the inclination angle is a, then Tan a = 2
The slope of M is Tan 2A = (2tan a) / (1 - (Tan a) ^ 2)=
-4/3
The intercept is 3, so the equation is 4x + 3y-9 = 0