If 5x + 6y - 34 = 0 and 2x + y + c = 0 are co | vedclass

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(C) The given circle is $x^2 + y^2 - 8x - 10y + 25 = 0$. Completing the square,we get $(x - 4)^2 + (y - 5)^2 = 16$. The center is $(4, 5)$ and the rad

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$(1, -5)$

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(C) The given circle is $x^2 + y^2 - 8x - 10y + 25 = 0$. Completing the square,we get $(x - 4)^2 + (y - 5)^2 = 16$. The center is $(4, 5)$ and the radius $r = 4$.Two lines $l_1x + m_1y + n_1 = 0$ and $l_2x + m_2y + n_2 = 0$ are conjugate with respect to the circle $(x - h)^2 + (y - k)^2 = r^2$ if $r^2(l_1l_2 + m_1m_2) = (l_1h + m_1k + n_1)(l_2h + m_2k + n_2)$.Here,$l_1 = 5, m_1 = 6, n_1 = -34$ and $l_2 = 2, m_2 = 1, n_2 = c$.Substituting the values: $16(5 	imes 2 + 6 	imes 1) = (5(4) + 6(5) - 34)(2(4) + 1(5) + c)$.$16(10 + 6) = (20 + 30 - 34)(8 + 5 + c)$.$16(16) = (16)(13 + c)$.$16 = 13 + c \implies c = 3$.The line is $2x + y + 3 = 0$.Checking the options,for $(1, -5)$: $2(1) + (-5) + 3 = 2 - 5 + 3 = 0$.Thus,the point $(1, -5)$ lies on the line.

If $5x + 6y - 34 = 0$ and $2x + y + c = 0$ are conjugate lines with respect to the circle $x^2 + y^2 - 8x - 10y + 25 = 0$,then which of the following points lies on the line $2x + y + c = 0$?

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