The value of k such that the circles x^2 + y^ | vedclass

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 + kx + 4y + 2 = 0$,we have $g_1 = \frac{k}{2}$

Vedclass · Accepted Answer

$\frac{-10}{3}$

Vedclass · Answer

(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 + kx + 4y + 2 = 0$,we have $g_1 = \frac{k}{2}$,$f_1 = 2$,and $c_1 = 2$.For the second circle,divide by $2$ to get $x^2 + y^2 - 2x - \frac{3}{2}y + \frac{k}{2} = 0$. Thus,$g_2 = -1$,$f_2 = -\frac{3}{4}$,and $c_2 = \frac{k}{2}$.The condition for two circles to intersect orthogonally is $2(g_1g_2 + f_1f_2) = c_1 + c_2$.Substituting the values: $2\left[\left(\frac{k}{2}\right)(-1) + (2)\left(-\frac{3}{4}\right)\right] = 2 + \frac{k}{2}$.$2\left[-\frac{k}{2} - \frac{3}{2}\right] = 2 + \frac{k}{2}$.$-k - 3 = 2 + \frac{k}{2}$.$-5 = k + \frac{k}{2} = \frac{3k}{2}$.$k = -\frac{10}{3}$.

The value of $k$ such that the circles $x^2 + y^2 + kx + 4y + 2 = 0$ and $2(x^2 + y^2) - 4x - 3y + k = 0$ cut orthogonally is:

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