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A rectangle $\mathrm{R}$ with end points of one of its sides as $(1,2)$ and $(3,6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2 x-y+4=0$ , then the area of $\mathrm{R}$ is______________.
Option: 1
16

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Let the cetre of the circle is $C(\alpha, 4+2 \alpha)$

It will be equidistant from $\mathrm{(1,2)(3,6)}$

$\mathrm{\Rightarrow(\alpha-1)^{2}+(4+2 \alpha-2)^{2}=(\alpha-3)^{2}+(4+2 \alpha-6)^{2}} \\$

$\mathrm{\Rightarrow \alpha^{2}-2 \alpha+1+4 \alpha^{2}+8 \alpha+4=\alpha^{2}-6 \alpha+3+4 \alpha^{2}-8 \alpha+4 }\\$

$\mathrm{\Rightarrow 20 \alpha=8 \Rightarrow \alpha=\frac{4}{5} ; \quad c:\left(\frac{4}{5}, \frac{28}{5}\right) .}$

$\mathrm{\text { distance b/w }(1,2) \quad \&(3,6)=2 \sqrt{5}=l e n g t h \text {. }}$

$\mathrm{Equation\: of\: line\: joining (1,2) and(3,6) is} \\$

$\mathrm{y-2=\frac{6-2}{3-1}(x-1)}$

$\mathrm{2x-y= 0}$

$\mathrm{distance\: of \: centre\: from \: 2 x-y=0\: is}\\$

$\mathrm{d=\frac{\left|2 \times \frac{4}{5}-\frac{28}{5}\right|}{\sqrt{2^{2}+1^{2}}}=\frac{4}{\sqrt{5}} \text { units } }$

$\mathrm{So\: Breadth\: \: 2 \times \frac{4}{\sqrt{5}}=\frac{8}{\sqrt{5}} units}\\$

$\mathrm{Area \: of \: R=l \times b=2 \sqrt{5} \times \frac{8}{\sqrt{5}}=16 \mathrm{sq} .u n i t s.}$

So answer is $\mathrm{16 \mathrm{sq} .u n i t s.}$

Posted by

Ritika Kankaria

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