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A square is inscribed in the circle x^2+y^2-10 x-6 y+30=0. One side of the square is parallel to y=x+3, then one vertex of the square is 

Option: 1

(3,3)


Option: 2

(7,3)


Option: 3

(6,3-\sqrt{3})


Option: 4

(6,3+\sqrt{3})


Answers (1)

best_answer

Let slope of OA is m, 

Then \left|\frac{m-1}{1+m}\right|=\tan 45^{\circ}

\Rightarrow \quad \frac{m-1}{1+m}= \pm 1

\begin{array}{llrl} \text { or } & & \frac{m-1}{m+1} & =-1 \\ \Rightarrow & & m-1 & =-m-1 \\ \therefore & m & =0 \end{array}

\\\therefore \text{Equation of} \mathrm{OA \: is} y=3\\ \\\text{Solving y=3 and} x^2+y^2-10 x-6 y+30=0

\begin{array}{lr} \Rightarrow \: \: \: \: \: \: \: \: \: \: x^2+9-10 x-18+30=0 \\ \Rightarrow \: \: \: \: \: \: \quad x^2-10 x+21=0 \\ \\ (x-7)(x-3)=0 \\ \\\text { or } \: \: \: x=3,7 \end{array}

\therefore  Two vertices are (3,3) and (7,3) and other diagonal is \perp to y=3 and through centre (5,3) is x=5

Now solving 

x=5 \text { and } x^2+y^2-10 x-6 y+30=0

\begin{array}{lr} \Rightarrow & 25+y^2-50-6 y+30=0 \\ \\\Rightarrow & y^2-6 y+5=0 \Rightarrow(y-1)(y-5)=0 \end{array}

\therefore \text { Other two vertices are }(5,1) \text { and }(5,5)

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