y = mx is a chord of a circle of radius a. Th | vedclass

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(B) The circle has radius $a$ and its diameter lies along the $x$-axis with one end at the origin $(0, 0)$. Thus,the center is $(a, 0)$ and the other

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$(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$

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(B) The circle has radius $a$ and its diameter lies along the $x$-axis with one end at the origin $(0, 0)$. Thus,the center is $(a, 0)$ and the other end of the diameter is $(2a, 0)$.The equation of the circle is $(x - a)^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0$.The chord $y = mx$ intersects the circle at the origin $O(0, 0)$ and point $B$. Substituting $y = mx$ into the circle equation:$x^2 + (mx)^2 - 2ax = 0$$x^2(1 + m^2) - 2ax = 0$$x(x(1 + m^2) - 2a) = 0$So,$x = 0$ or $x = \frac{2a}{1 + m^2}$.For $x = \frac{2a}{1 + m^2}$,$y = m \left( \frac{2a}{1 + m^2} \right) = \frac{2am}{1 + m^2}$.Thus,$B = \left( \frac{2a}{1 + m^2}, \frac{2am}{1 + m^2} \right)$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Using $O(0, 0)$ and $B\left( \frac{2a}{1 + m^2}, \frac{2am}{1 + m^2} \right)$:$x\left( x - \frac{2a}{1 + m^2} \right) + y\left( y - \frac{2am}{1 + m^2} \right) = 0$$x^2 + y^2 - \frac{2ax}{1 + m^2} - \frac{2amy}{1 + m^2} = 0$$(1 + m^2)(x^2 + y^2) - 2a(x + my) = 0$.

$y = mx$ is a chord of a circle of radius $a$. The diameter of the circle lies along the $x$-axis and one end of this chord is at the origin. The equation of the circle described on this chord as diameter is

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