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

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(D) Let the pole be $(h, k)$.The equation of the polar of the point $(h, k)$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is given by $T

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$(3, -1)$

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(D) Let the pole be $(h, k)$.The equation of the polar of the point $(h, k)$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is given by $T = 0$.Dividing the circle equation by $2$,we get $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$.The equation of the polar is $hx + ky - \frac{3}{2}(\frac{x+h}{2}) + \frac{5}{2}(\frac{y+k}{2}) - \frac{7}{2} = 0$.Multiplying by $4$,we get $4hx + 4ky - 3(x + h) + 5(y + k) - 14 = 0$.$(4h - 3)x + (4k + 5)y - 3h + 5k - 14 = 0$.Comparing this with the given line $9x + y - 28 = 0$,we have:$\frac{4h - 3}{9} = \frac{4k + 5}{1} = \frac{-3h + 5k - 14}{-28} = \lambda$.From $\frac{4h - 3}{9} = \lambda$,$4h - 3 = 9\lambda \Rightarrow 4h = 9\lambda + 3$.From $\frac{4k + 5}{1} = \lambda$,$4k + 5 = \lambda \Rightarrow 4k = \lambda - 5$.Substituting into the third ratio: $\frac{-3h + 5k - 14}{-28} = \lambda \Rightarrow -3h + 5k - 14 = -28\lambda$.Multiplying by $4$: $-3(4h) + 5(4k) - 56 = -112\lambda$.$-3(9\lambda + 3) + 5(\lambda - 5) - 56 = -112\lambda$.$-27\lambda - 9 + 5\lambda - 25 - 56 = -112\lambda$.$-22\lambda - 90 = -112\lambda$ $\Rightarrow 90\lambda = 90$ $\Rightarrow \lambda = 1$.Thus,$4h = 9(1) + 3 = 12 \Rightarrow h = 3$.$4k = 1 - 5 = -4 \Rightarrow k = -1$.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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