The pole of the straight line 9x + y - 28 = 0 | vedclass

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(C) Let $(h, k)$ be the pole of the line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$. The equation of the polar of a

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$(3, -1)$

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(C) Let $(h, k)$ be the pole of the line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$. The equation of the polar of a point $(h, k)$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is given by $T = 0$. The equation is $2hx + 2ky - \frac{3(x + h)}{2} + \frac{5(y + k)}{2} - 7 = 0$. Multiplying by $2$,we get $4hx + 4ky - 3x - 3h + 5y + 5k - 14 = 0$. Rearranging terms,we get $(4h - 3)x + (4k + 5)y + (5k - 3h - 14) = 0$. Comparing this with the given line $9x + y - 28 = 0$,we have: $\frac{4h - 3}{9} = \frac{4k + 5}{1} = \frac{5k - 3h - 14}{-28}$. From $\frac{4h - 3}{9} = 4k + 5$,we get $4h - 3 = 36k + 45$ $\Rightarrow 4h - 36k = 48$ $\Rightarrow h - 9k = 12$. From $\frac{4k + 5}{1} = \frac{5k - 3h - 14}{-28}$,we get $-112k - 140 = 5k - 3h - 14$ $\Rightarrow 3h - 117k = 126$ $\Rightarrow h - 39k = 42$. Solving $h - 9k = 12$ and $h - 39k = 42$,we subtract the equations: $(h - 9k) - (h - 39k) = 12 - 42$ $\Rightarrow 30k = -30$ $\Rightarrow k = -1$. Substituting $k = -1$ into $h - 9k = 12$,we get $h - 9(-1) = 12$ $\Rightarrow h + 9 = 12$ $\Rightarrow h = 3$. Thus,the pole is $(3, -1)$.

The pole of the straight line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is

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