Through the point of intersection $P$ of the circle $x^2 + y^2 = 1$ and $x^2 + y^2 = 2x + 4y + 1 = 0$ a common chord $APB$ is drawn terminating on the two circles such that the chords $AP$ and $BP$ of the given circles subtend equal angles at the respective centres. If the coordinates of $P$ are integral and the equation of the chord is $y = 2mx + 1$ then the value of $m$ is ___.
Step-by-Step Solution
Key Concept: Find intersection points by subtracting circle equations to get the radical axis, then solve simultaneously.
Solve $x^2 + y^2 = 1$ and $x^2 + y^2 + 2x + 4y + 1 = 0$ simultaneously. From the first equation substitute into the second to get $2x + 4y + 1 = 0$, which gives $y = 0$ or $y = -\frac{4}{5}$, $x = -1$ or $x = \frac{3}{5}$. The circles intersect at $(-1, 0)$ and $(\frac{3}{5}, -\frac{4}{5})$. For $P = (-1, 0)$ to have integer coordinates, the slope is $m = \frac{0 - (-1)}{-1 - (-1)}$, undefined. Since $P$ is $(-1, 0)$, we get $0 = -2m + 1$, so $m = 0.5$.
Correct Answer: 0.5