Question

Locus of the intersection of the two straight lines passing through (1, 0) and (-1, 0) respectively and including an angle of 45° can be a circle with

• A. centre (1, 0) and radius $\sqrt{2}$.
• B. centre (1, 0) and radius 2.
• C. centre (0, 1) and radius $\sqrt{2}$.
• D. centre (0, -1) and radius $\sqrt{2}$.

Answer 1

Answer: (C), (D)

C, D

Answer 2

Let the two points be $A = (1, 0)$ and $B = (-1, 0)$. Let $P = (x, y)$ be the point of intersection. The lines passing through $A$ and $B$ have slopes $m_1 = \frac{y}{x-1}$ and $m_2 = \frac{y}{x+1}$ respectively. The angle $\theta$ between them is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$. Given $\theta = 45^\circ$, $\tan 45^\circ = 1$. Substituting the slopes: $1 = \left| \frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + \frac{y}{x - 1} \cdot \frac{y}{x + 1}} \right| = \left| \frac{2y}{x^2 - 1 + y^2} \right|$ This leads to two cases:

1. $\frac{2y}{x^2 - 1 + y^2} = 1 \implies x^2 + y^2 - 2y - 1 = 0 \implies x^2 + (y-1)^2 = 2$. This is a circle with center $(0, 1)$ and radius $\sqrt{2}$.
2. $\frac{2y}{x^2 - 1 + y^2} = -1 \implies x^2 + y^2 + 2y - 1 = 0 \implies x^2 + (y+1)^2 = 2$. This is a circle with center $(0, -1)$ and radius $\sqrt{2}$. The locus is the union of these two circles.