The two circles x^2 + y^2 - 2x + 22y + 5 = 0 | vedclass

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(A) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 - 2x + 22y + 5 = 0$,we have $g_1 = -1$,$f_1 =

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$47$

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(A) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 - 2x + 22y + 5 = 0$,we have $g_1 = -1$,$f_1 = 11$,and $c_1 = 5$.For the second circle $x^2 + y^2 + 14x + 6y + k = 0$,we have $g_2 = 7$,$f_2 = 3$,and $c_2 = k$.Two circles intersect orthogonally if $2(g_1g_2 + f_1f_2) = c_1 + c_2$.Substituting the values,we get $2((-1)(7) + (11)(3)) = 5 + k$.$2(-7 + 33) = 5 + k$.$2(26) = 5 + k$.$52 = 5 + k$.$k = 52 - 5 = 47$.

The two circles $x^2 + y^2 - 2x + 22y + 5 = 0$ and $x^2 + y^2 + 14x + 6y + k = 0$ intersect orthogonally provided $k$ is equal to

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