Question

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Answer 1

Topic :-

$\mathtt{Circle}$

Given :-

$\mathtt{The\:line\:\bold{y - x + 3 = 0}\: is\: normal\:to\:a\:circle\:at\: point\:\left(3+\dfrac{3}{\sqrt{2}},\dfrac{3}{\sqrt{2}} \right).}$

To Find :-

$\mathtt{Equation\:of\:Circle\:among\:given\:options\:which\:satisfies\:given\:condition.}$

Concept Used :-

$\mathtt{If\:(h, k)\:is\:centre\:of\:the\:circle\:then\:equation\:of\:circle\:is\:given\:by:-}$

$\mathtt{(x-h)^2+(y-k)^2=r^2}$

$\mathtt{where\:r\:is\:radius\:of\:Circle.}$

$\mathtt{Normal\:of\:the\:circle\:passes\:through\:its\:centre.}$

Solution :-

$\mathtt{Normal\:of\:the\:circle\:passes\:through\:its\:centre.}$

$\mathtt{So,centre\:of\:the\:circle\:will\:satisfy\:the\:equation\:of\:Normal.}$

$\mathtt{So,y-x+3=0\:passes\:through\:(h,k),which\:means,}$

$\mathtt{k-h+3=0}$

$\bold{Now,we\:will\:be\:checking\:all\:options,}$

$\bold{option\:which\:will\:satisfy\:k-h+3=0\:will\:be\:correct\:option.}$

Checking option (A),

$\mathtt{\left ( x-3-\dfrac{3}{\sqrt{2}}\right )^2+\left ( y-\dfrac{3}{\sqrt{2}} \right )^2=9}$

$\mathtt{On\:comparing\:it\:with\:equation\:of\:circle,we\:get,}$

$\mathtt{h=3+\dfrac{3}{\sqrt{2}}}$

$\mathtt{k=\dfrac{3}{\sqrt{2}}}$

$\mathtt{This\:is\:the\:point\:from\:which\:we\:are\:drawing\:normal\:to\:Circle.}$

$\mathtt{Hence,this\:is\:wrong\:option.}$

Checking option (B),

$\mathtt{\left ( x-\dfrac{3}{\sqrt{2}}\right )^2+\left ( y-\dfrac{3}{\sqrt{2}} \right )^2=9}$

$\mathtt{On\:comparing\:it\:with\:equation\:of\:circle,we\:get,}$

$\mathtt{h=\dfrac{3}{\sqrt{2}}}$

$\mathtt{k=\dfrac{3}{\sqrt{2}}}$

$\mathtt {k-h+3\neq 0}$

$\mathtt{Hence,this\:is\:wrong\:option.}$

Checking option (C),

$\mathtt{x^2+(y-3)^2=9}$

$\mathtt{On\:comparing\:it\:with\:equation\:of\:circle,we\:get,}$

$\mathtt{h=0}$

$\mathtt{k=3}$

$\mathtt {k-h+3\neq 0}$

$\mathtt{Hence,this\:is\:wrong\:option.}$

Checking option (D),

$\mathtt{(x-3)^2+y^2=9}$

$\mathtt{On\:comparing\:it\:with\:equation\:of\:circle,we\:get,}$

$\mathtt{h=3}$

$\mathtt{k=0}$

$\mathtt {k-h+3= 0}$

$\mathtt{Hence,this\:is\:correct\:option.}$

Answer :-

$\mathtt{Hence,\bold{option\:D:}\:(x-3)^2+y^2=9,\:is\:correct\:option.}$