If a variable circle S=0 touches the line y=x | vedclass

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(A) Let the equation of the variable circle $S=0$ passing through the origin $(0,0)$ be $x^2+y^2+2gx+2fy=0$.Since the circle touches the line $x-y=0$,

Vedclass · Accepted Answer

$\left(\frac{1}{2}, \frac{1}{2}\right)$

Vedclass · Answer

(A) Let the equation of the variable circle $S=0$ passing through the origin $(0,0)$ be $x^2+y^2+2gx+2fy=0$.Since the circle touches the line $x-y=0$,the perpendicular distance from the center $(-g, -f)$ to the line must equal the radius $\sqrt{g^2+f^2}$.$\frac{|-g+f|}{\sqrt{1^2+(-1)^2}} = \sqrt{g^2+f^2} \Rightarrow \frac{(g-f)^2}{2} = g^2+f^2$.$g^2+f^2-2gf = 2g^2+2f^2$ $\Rightarrow g^2+f^2+2gf = 0$ $\Rightarrow (g+f)^2 = 0$ $\Rightarrow f = -g$.Substituting $f = -g$ into the circle equation,we get $S = x^2+y^2+2gx-2gy = 0$.The common chord of $S=0$ and $x^2+y^2+6x+8y-7=0$ is given by $S_1 - S_2 = 0$.$(x^2+y^2+2gx-2gy) - (x^2+y^2+6x+8y-7) = 0$.$(2g-6)x - (2g+8)y + 7 = 0$.$2g(x-y) - (6x+8y-7) = 0$.This represents a family of lines passing through the intersection of $x-y=0$ and $6x+8y-7=0$.Solving these: $x=y$,so $6x+8x-7=0$ $\Rightarrow 14x=7$ $\Rightarrow x=\frac{1}{2}$.Thus,the fixed point is $\left(\frac{1}{2}, \frac{1}{2}\right)$.

If a variable circle $S=0$ touches the line $y=x$ and passes through the point $(0,0)$,then the fixed point that lies on the common chord of the circles $x^2+y^2+6x+8y-7=0$ and $S=0$ is

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