Question

Find the equation of the chord of $x^2 + y^2 - 6x + 10 - a = 0$ which is bisected at $(-2, 4)$.

Answer 1

The equation of the chord is $5x - 4y + 26 = 0$.

Answer 2

The equation of a chord of a circle $S=0$ bisected at a point $(x_1, y_1)$ is given by $T = S_1$. Here, $S \equiv x^2 + y^2 - 6x + 10 - a = 0$ and the point of bisection is $(x_1, y_1) = (-2, 4)$.

$S_1$ is the value of $S$ at $(-2, 4)$: $S_1 = (-2)^2 + (4)^2 - 6(-2) + 10 - a$ $S_1 = 4 + 16 + 12 + 10 - a$ $S_1 = 42 - a$

$T$ is obtained by replacing $x^2$ with $xx_1$, $y^2$ with $yy_1$, $x$ with $\frac{x+x_1}{2}$, and $y$ with $\frac{y+y_1}{2}$: $T = x(-2) + y(4) - 6\left(\frac{x+(-2)}{2}\right) + 10 - a$ $T = -2x + 4y - 3(x-2) + 10 - a$ $T = -2x + 4y - 3x + 6 + 10 - a$ $T = -5x + 4y + 16 - a$

The equation of the chord is $T = S_1$: $-5x + 4y + 16 - a = 42 - a$ $-5x + 4y + 16 = 42$ $-5x + 4y = 42 - 16$ $-5x + 4y = 26$ $5x - 4y + 26 = 0$