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(B) Given the equation of the line $L$ is $x-y=0$. The distance $b$ from a variable point $P(\alpha, \beta)$ to the line $x-y=0$ is given by $b = \lef

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$3x^2+3y^2-10xy+8x+4y-10=0$

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(B) Given the equation of the line $L$ is $x-y=0$. The distance $b$ from a variable point $P(\alpha, \beta)$ to the line $x-y=0$ is given by $b = \left|\frac{\alpha-\beta}{\sqrt{2}}ight|$,which implies $b^2 = \frac{(\alpha-\beta)^2}{2}$,or $(\alpha-\beta)^2 = 2b^2$ $\ldots(i)$. The distance $a$ between $P(\alpha, \beta)$ and $A(2,1)$ is given by $a^2 = (\alpha-2)^2 + (\beta-1)^2$ $\ldots(ii)$. The distance $c$ from the origin $(0,0)$ to $A(2,1)$ is $c = \sqrt{2^2+1^2} = \sqrt{5}$,so $c^2 = 5$. Given the condition $a = bc$,we have $a^2 = b^2c^2 = 5b^2$ $\ldots(iii)$. Substituting $(i)$ and $(ii)$ into $(iii)$,we get $(\alpha-2)^2 + (\beta-1)^2 = 5 	imes \frac{(\alpha-\beta)^2}{2}$. Multiplying by $2$,we get $2(\alpha^2 - 4\alpha + 4 + \beta^2 - 2\beta + 1) = 5(\alpha^2 + \beta^2 - 2\alpha\beta)$. $2\alpha^2 - 8\alpha + 8 + 2\beta^2 - 4\beta + 2 = 5\alpha^2 + 5\beta^2 - 10\alpha\beta$. Rearranging the terms,we get $3\alpha^2 + 3\beta^2 - 10\alpha\beta + 8\alpha + 4\beta - 10 = 0$. Replacing $(\alpha, \beta)$ with $(x, y)$,the locus of $P$ is $3x^2 + 3y^2 - 10xy + 8x + 4y - 10 = 0$.

Let $A(2,1)$ be a point and the equation of the straight line $L$ be $x-y=0$. Let $a$ and $b$ respectively represent the distances from a variable point $P(\alpha, \beta)$ to $A$ and to the line $L$. If $c$ is the distance of the point $A$ from the origin such that $a=bc$,then the locus of $P$ is

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