Explanation
The image of circle w.r.t same line means image of centre w.r.t that line without changing radius $$x^{2}+y^{2}-6x-4y+12=0$$ Centre $$=(3,2)$$ Radius$$=1$$ Image of $$(3,2)$$ w.r.t $$x+y-1=0$$ $$\dfrac {x-3}{1}=\dfrac {y-2}{1}=-2\dfrac {(3+2-1)}{1^{1}+1^{2}}=-4$$ $$x=-1$$, $$y=-2$$ Then equation of image of circle is $$(x+1)^{2}+(y+2)^{2}=(1)^{2}$$ $$\Rightarrow x^{2}+y^{2}+2x+4y+4=0$$
Given equation is $$x^{2}+y^{2}=2ax$$
Centre $$=(a,0)$$ and radius $$=a$$
Therefore, equation of line having slope $$=$$ $$\dfrac {-1}{2}$$ and passing through $$(4,0)$$ $$y=\dfrac {-1}{2}(x-a)$$ Line $$AB$$: $$2y+x=a$$ -(1) $$y=\dfrac {a+x}{2}$$ $$d=\left | \dfrac {a}{\sqrt{5}} \right |$$ So, area of $$\triangle AOB =\dfrac {1}{2}\times\ AB\ \times$$ (perpendicular distance of $$Q$$ from $$AB$$ ) $$=\dfrac {1}{2}\times 2a\times \dfrac {a}{\sqrt{5}}$$ $$=\dfrac {a^{2}}{\sqrt{5}}$$
$$d=\left | \dfrac{C_{1}-C_{2}}{\sqrt{a^{2}+b^{2}}} \right |$$where $$d$$ is the distance between lines whose equations are $$ax+by+C_{1}=0$$ & $$ax+by+C_{2}=0$$$$d=\left | \dfrac{27-2}{\sqrt{4^{2}+3^2}} \right |$$$$=5$$$$d=5$$If the distance between vertex and latus rectum$$=$$distance of vertex from directrix$$=a$$
then $$d=2a=5$$$$\Rightarrow$$ Length of latus rectum$$=4a=10$$
$${\textbf{Step - 1: Find value of 'a' from Length of latus rectum,}}$$
$${\text{We know that length of latus rectum of any parabola is 4a.}}$$
$${\text{Given that 4a = 2 }} \Rightarrow {\text{a = }}\dfrac{1}{2}.$$
$${\text{Also given that the axis is x + y = 2}}.$$
$${\textbf{Step - 2: Use relation between Perpendicular distance from axis and Distance from tangent of a parabola.}}$$
$${\text{From parabola's formula we know that,}}$$
$$\Rightarrow {\left( {{\text{perpendicular distance from axis}}} \right)^2}{\text{ = 4a}}\left( {{\text{distance from tangent}}} \right).$$
$$ \Rightarrow {\text{ }}{\left( {\dfrac{{{\text{x + y}} - {\text{2}}}}{{\sqrt 2 }}} \right)^2} = 4\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{{\text{x}} - {\text{y + 4}}}}{{\sqrt 2 }}} \right).$$
$$ \Rightarrow {\left( {{\text{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\text{x}} - {\text{y + 4}}} \right).$$
$${\textbf{Hence, The equation is }}{\left( {{\textbf{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\textbf{x}} - {\textbf{y + 4}}} \right).(C)$$
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