Question

If y = mx be the equation of chord of a circle whose radius is a. Origin being one extremity of the chord and the axis of x being a diameter of the circle, prove that the equation of a circle of which this chord is the diameter is $(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$

• A. The equation of the circle is $(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$.
• B. The equation of the circle is $(1 + m^2)(x^2 + y^2) + 2a(x + my) = 0$.
• C. The equation of the circle is $(1 - m^2)(x^2 + y^2) - 2a(x + my) = 0$.
• D. The equation of the circle is $(1 + m^2)(x^2 + y^2) - 2a(x - my) = 0$.

Answer 1

Answer: (A)

The equation of the circle is $(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$.

Answer 2

1. Define Circle 1: The problem states that the x-axis is a diameter of a circle with radius $a$, and the origin (0,0) is one extremity of a chord. This implies the origin is a point on the circle. If the x-axis is a diameter and the origin is on the circle, the diameter along the x-axis must have one endpoint at the origin. This diameter can be from (0,0) to (2a,0) or from (0,0) to (-2a,0).

   • If the diameter is (0,0) to (2a,0), the center of Circle 1 is $(a,0)$ and its equation is $(x-a)^2 + y^2 = a^2$.
   • If the diameter is (0,0) to (-2a,0), the center of Circle 1 is $(-a,0)$ and its equation is $(x+a)^2 + y^2 = a^2$.

2. Find the Chord's Endpoints: The chord has the equation $y=mx$ and one extremity is the origin O(0,0). Let the other extremity be B$(x_1, y_1)$. Since B lies on $y=mx$, $y_1 = mx_1$. B must also lie on Circle 1.

   • Case 1: Circle 1 is $(x-a)^2 + y^2 = a^2$. Substituting $y_1=mx_1$: $(x_1-a)^2 + (mx_1)^2 = a^2$ $x_1^2 - 2ax_1 + a^2 + m^2x_1^2 = a^2$ $x_1^2(1+m^2) - 2ax_1 = 0$ $x_1(x_1(1+m^2) - 2a) = 0$ The solutions are $x_1=0$ (which is the origin O) and $x_1 = \frac{2a}{1+m^2}$. For $x_1 = \frac{2a}{1+m^2}$, $y_1 = mx_1 = \frac{2am}{1+m^2}$. So, the other extremity of the chord is $B = (\frac{2a}{1+m^2}, \frac{2am}{1+m^2})$.
   • Case 2: Circle 1 is $(x+a)^2 + y^2 = a^2$. Substituting $y_1=mx_1$: $(x_1+a)^2 + (mx_1)^2 = a^2$ $x_1^2 + 2ax_1 + a^2 + m^2x_1^2 = a^2$ $x_1^2(1+m^2) + 2ax_1 = 0$ $x_1(x_1(1+m^2) + 2a) = 0$ The solutions are $x_1=0$ and $x_1 = \frac{-2a}{1+m^2}$. For $x_1 = \frac{-2a}{1+m^2}$, $y_1 = mx_1 = \frac{-2am}{1+m^2}$. So, the other extremity of the chord is $B = (\frac{-2a}{1+m^2}, \frac{-2am}{1+m^2})$.

3. Define Circle 2: The chord $y=mx$ is the diameter of a new circle (Circle 2). The endpoints of this diameter are O(0,0) and B$(x_1, y_1)$. The equation of a circle with diameter endpoints $(x_A, y_A)$ and $(x_B, y_B)$ is $(x-x_A)(x-x_B) + (y-y_A)(y-y_B) = 0$. With O(0,0) as one endpoint, the equation is $x(x-x_1) + y(y-y_1) = 0$.

4. Derive the Equation for Circle 2: The problem asks to prove a specific equation: $(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$. This equation has a $-2a$ term.

   • If we use the endpoint B from Case 1: $B = (\frac{2a}{1+m^2}, \frac{2am}{1+m^2})$. The equation of Circle 2 is: $x(x - \frac{2a}{1+m^2}) + y(y - \frac{2am}{1+m^2}) = 0$ $x^2 - \frac{2ax}{1+m^2} + y^2 - \frac{2amy}{1+m^2} = 0$ Multiply by $(1+m^2)$: $(1+m^2)(x^2 + y^2) - 2ax - 2amy = 0$ $(1+m^2)(x^2 + y^2) - 2a(x + my) = 0$. This matches the equation to be proven.

   • If we use the endpoint B from Case 2: $B = (\frac{-2a}{1+m^2}, \frac{-2am}{1+m^2})$. The equation of Circle 2 is: $x(x - (\frac{-2a}{1+m^2})) + y(y - (\frac{-2am}{1+m^2})) = 0$ $x(x + \frac{2a}{1+m^2}) + y(y + \frac{2am}{1+m^2}) = 0$ $(1+m^2)(x^2 + y^2) + 2a(x + my) = 0$. This does not match the equation to be proven.

   Therefore, the problem implicitly assumes the configuration of Circle 1 that leads to the specific equation required. This configuration is when the diameter on the x-axis extends from (0,0) to (2a,0).