If OA and OB are two equal chords of the circle $x^2+y^2-2x+4y=0$ perpendicular to each other and passing through the origin, the slopes of OA and OB are the roots of the equation.
- $3m^2+8m-3=0$
- $3m^2-8m-3=0$
- $8m^2+3m-8=0$
- $8m^2+3m+8=0$
The circle x^2 + y^2 - 2x + 4y = 0 has center (1, -2). Chords OA and OB pass through origin (0,0). Since they are perpendicular, their slopes m1 and m2 satisfy m1 * m2 = -1. The equation of the lines is y = mx. Substituting into the circle equation: x^2 + (mx)^2 - 2x + 4(mx) = 0 -> x^2(1+m^2) - 2x(1-2m) = 0. For the chords, x(1+m^2) = 2(1-2m). This leads to the quadratic in m.
The given circle equation x squared plus y squared minus 2x plus 4y equals zero has a center at (1, -2). A line passing through the origin with slope m has the equation y equals m times x, which can be written as m times x minus y equals 0. The length of a chord drawn from a point on the circle to another point is found using the perpendicular distance from the center to the line, so the distance from (1, -2) to the line m times x minus y equals 0 is the square root of (m plus 2) squared divided by (m squared plus 1). Since the two chords are perpendicular, their slopes m and -1 divided by m must yield equal chord lengths. Setting the squared distances equal gives ((m plus 2) squared) divided by (m squared plus 1) equals ((-1 divided by m plus 2) squared) divided by ((1 divided by m squared) plus 1). Solving this proportion yields 3 times m squared minus 8 times m minus 3 equals 0.