Equation of the chord of the circle $x^{2} + y^{2} + 4x - 6y - 9 = 0$ bisected at $(0, 1)$ is
- $y - 1 = x$
- $y + 1 = x$
- $y + 1 = 2x$
- $y - 1 = 3x$
The equation of a chord of a circle bisected at (x1, y1) is T = S1. Here, T = x(0) + y(1) + 2(x+0) - 3(y+1) - 9 = y + 2x - 3y - 3 - 9 = 2x - 2y - 12. S1 = 0^2 + 1^2 + 4(0) - 6(1) - 9 = 1 - 6 - 9 = -14. Setting 2x - 2y - 12 = -14 gives 2x - 2y = -2, or x - y = -1, which simplifies to y - 1 = x.
Use the formula for a chord bisected at a given point, T equals S1, where T is x times x1 plus y times y1 plus 2 times x plus x1 minus 3 times y plus y1 minus 9 and S1 is 0 squared plus 1 squared plus 4 times 0 minus 6 times 1 minus 9. This simplifies to x times 0 plus y times 1 plus 2 times x minus 3 times y plus 3 minus 9 equals -14, or 2x minus 2y plus 2 equals 0. Dividing the entire equation by 2 and rearranging the terms yields y minus 1 equals x.