The locus of the centre of a variable circle | vedclass

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(D) Let the equation of the variable circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2

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$x - y = 0$

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(D) Let the equation of the variable circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.For the first circle $x^2 + y^2 - 2x - 4y - 1 = 0$,we have $g_1 = -1, f_1 = -2, c_1 = -1$. The condition is $2g(-1) + 2f(-2) = c - 1$,which simplifies to $-2g - 4f = c - 1$  .......$(1)$For the second circle $x^2 + y^2 - 4x - 2y - 1 = 0$,we have $g_2 = -2, f_2 = -1, c_2 = -1$. The condition is $2g(-2) + 2f(-1) = c - 1$,which simplifies to $-4g - 2f = c - 1$  .......$(2)$Subtracting equation $(2)$ from equation $(1)$:$(-2g - 4f) - (-4g - 2f) = (c - 1) - (c - 1)$$2g - 2f = 0$$g - f = 0$Since the centre of the variable circle is $(-g, -f)$,let the centre be $(x, y)$. Thus $x = -g$ and $y = -f$.Substituting $g = -x$ and $f = -y$ into $g - f = 0$,we get $-x - (-y) = 0$,which simplifies to $x - y = 0$.

The locus of the centre of a variable circle which cuts the circles $x^2 + y^2 - 2x - 4y - 1 = 0$ and $x^2 + y^2 - 4x - 2y - 1 = 0$ orthogonally is:

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