The coordinates of the mid-point of the chord | vedclass

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(A) Let the circle be $x^{2} + y^{2} - 6x + 2y - 54 = 0$. The center $O$ is $(3, -1)$.Let $M(h, k)$ be the mid-point of the chord.Since $OM$ is perpen

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$(1, 4)$

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(A) Let the circle be $x^{2} + y^{2} - 6x + 2y - 54 = 0$. The center $O$ is $(3, -1)$.Let $M(h, k)$ be the mid-point of the chord.Since $OM$ is perpendicular to the chord $2x - 5y + 18 = 0$,the slope of $OM$ is the negative reciprocal of the slope of the line.The slope of the line $2x - 5y + 18 = 0$ is $m = \frac{2}{5}$.Thus,the slope of $OM$ is $-\frac{5}{2}$.The slope of $OM$ is also given by $\frac{k - (-1)}{h - 3} = \frac{k + 1}{h - 3}$.Equating the slopes: $\frac{k + 1}{h - 3} = -\frac{5}{2}$ $\Rightarrow 2k + 2 = -5h + 15$ $\Rightarrow 5h + 2k = 13$.Since $M(h, k)$ lies on the line $2x - 5y + 18 = 0$,we have $2h - 5k = -18$.Solving the system:$5h + 2k = 13$ (multiply by $5$): $25h + 10k = 65$$2h - 5k = -18$ (multiply by $2$): $4h - 10k = -36$Adding the equations: $29h = 29 \Rightarrow h = 1$.Substituting $h = 1$ into $2h - 5k = -18$: $2(1) - 5k = -18$ $\Rightarrow -5k = -20$ $\Rightarrow k = 4$.The mid-point is $(1, 4)$.

The coordinates of the mid-point of the chord cut off by the line $2x - 5y + 18 = 0$ by the circle $x^{2} + y^{2} - 6x + 2y - 54 = 0$ are:

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