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Line y = x touches the circle at P such that \mathrm{O P=4 \sqrt{2}}  (where O is the origin). Line x + y = 0 contains the chord of the circle having length of  \mathrm{4 \sqrt{2}}\: unit Point (2, 8) lies inside the circle. Find the equation of the circle.


 

Option: 1

\mathrm{(x+1)^{2}+(y-a)^{2}=50}


Option: 2

\mathrm{(x-a)^{2}+(y-1)^{2}=50}


Option: 3

\mathrm{(x+1)^{2}+(y-a)^{2}=25}


Option: 4

None of these


Answers (1)

best_answer

OPSR  is rectangle 

\begin{aligned} & \mathrm{OP}=4 \sqrt{2} \Rightarrow \mathrm{SR}=4 \sqrt{2} \\ & \mathrm{QR}=\frac{6 \sqrt{3}}{2}=3 \sqrt{3} \end{aligned}

∴ SQ = radius of the circle 

\mathrm{r=5 \sqrt{2}}

y = x touches the circle 

\mathrm{\begin{aligned} \frac{h-k}{\sqrt{2}} & = \pm 5 \sqrt{2} \\ \therefore h-k & = \pm 10 \end{aligned}}--------(1)

And \mathrm{\mathrm{SR}=4 \sqrt{2} \Rightarrow \frac{\mathrm{h}+\mathrm{k}}{\sqrt{2}}= \pm 4 \sqrt{2} \Rightarrow h+k= \pm 8}----(2)

From (1) and (2) centre of circle may be (9, -1), (1, -9), (-1, 9) and (-9, 1) 

Hence  equation of the circles are
\mathrm{\begin{aligned} & (x-9)^2+(y+1)^2=50 \\ & (x-1)^2+(y+9)^2=50 \\ & (x+1)^2+(y-9)^2=50 \\ & (x+9)^2+(y-1)^2=50 \end{aligned}}

Now (2, 8) lies inside the circle. 

LHS < 50 for (3). Hence answer is equation (3).

 

 

 

 

Posted by

Gautam harsolia

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