$x^{2}+y^{2}-\frac{(1+a) x}{2}-\frac{(1-a) y}{2}=0$

Centre $\left(\frac{1+\mathrm{a}}{4}, \frac{1-\mathrm{a}}{4}\right) \Rightarrow(\mathrm{h}, \mathrm{k})$

$\mathrm{P}\left(\frac{1+\mathrm{a}}{2}, \frac{1-\mathrm{a}}{2}\right) \Rightarrow(2 \mathrm{h}, 2 \mathrm{k})$

JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Circle Question 64 English Explanation

JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Circle Question 64 English Explanation

Equation of chord $\Rightarrow \mathrm{T}=\mathrm{S}_{1}$

$\Rightarrow(\mathrm{x}-\mathrm{y}) \lambda-\frac{2 \mathrm{~h}(\mathrm{x}+\lambda)}{2}-\frac{(2 \mathrm{k})(\mathrm{y}-\lambda)}{2}$

$=2 \lambda^{2}-2 \mathrm{~h}(\lambda)+2 \mathrm{k} \lambda$

Now, $\lambda(2 \mathrm{~h}, 2 \mathrm{k})$ satisfies the chord

$\therefore(2 \mathrm{~h}-2 \mathrm{k}) \lambda-\mathrm{h}(\mathrm{x}+\lambda)-\mathrm{k}(\mathrm{y}-\lambda)$

$\Rightarrow 2 \lambda^{2}+4 \mathrm{k} \lambda-4 \mathrm{~h} \lambda+\mathrm{h} \lambda-\mathrm{k} \lambda+\mathrm{hx}+\mathrm{ky}=0$

$\Rightarrow 2 \lambda^{2}+\lambda(3 \mathrm{k}-3 \mathrm{~h})+\mathrm{ky}+\mathrm{hx}=0$

$\Rightarrow \mathrm{D}>0$

$\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8(\mathrm{ky}+\mathrm{hx})>0$

$\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8\left(2 \mathrm{k}^{2}+2 \mathrm{~h}^{2}\right)>0$

$\Rightarrow-7 \mathrm{k}^{2}-7 \mathrm{~h}^{2}-18 \mathrm{kh}>0$

$\Rightarrow 7 \mathrm{k}^{2}+7 \mathrm{~h}^{2}+18 \mathrm{kh}<0$

$\Rightarrow 7\left(\frac{1-\mathrm{a}}{4}\right)^{2}+7\left(\frac{1+\mathrm{a}}{4}\right)^{2}+18\left(\frac{1-\mathrm{a}^{2}}{16}\right)<0$

$\Rightarrow 7\left[\frac{2\left(1+\mathrm{a}^{2}\right)}{16}\right]+\frac{18\left(1-\mathrm{a}^{2}\right)}{16}<0, \quad \mathrm{a}^{2}=\mathrm{t}$

$\Rightarrow \frac{7}{8}(1+\mathrm{t})+\frac{18(1-\mathrm{t})}{16}<0$

$\Rightarrow \frac{14+14 \mathrm{t}+18-18 \mathrm{t}}{16}<0$

$\Rightarrow 4 \mathrm{t}>32$

$$ \Rightarrow $$ $\mathrm{t}>8 $

$$ \Rightarrow $$ $ \mathrm{a}^{2}>8$

Circle

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