CGPET 2015 Solved Paper

Let required point be (α,β) on the straight line $y=2x+11$, which is nearest to the circle $16(x^2+y^2)+32x-8y-50=0$ ⇒ $x^2+y^2+2x-\frac{1}{2}y-\frac{50}{16}=0$ ∴ centre of circle $=\left(-1,\frac{1}{4}\right)$ and radius, $r=\sqrt{1+\frac{1}{16}+\frac{50}{16}}=\frac{\sqrt{67}}{4}$ Now, equation of straight line passing through centre $\left(-1,\frac{1}{4}\right)$ and (α,β) is $y-\frac{1}{4}=\frac{\beta-\frac{1}{4}}{\alpha+1}(x+1)$ ...(i) Now, gradient of this straight line $=\frac{\beta-\frac{1}{4}}{\alpha+1}$ Since, straight line (i) is perpendicular to the line $y=2x+11$ ∴ $\frac{\beta-\frac{1}{4}}{\alpha+1} \times 2 = -1 \quad [\because m_1 \cdot m_2 = -1]$ ⇒ $2\beta - \frac{1}{2} = -\alpha - 1$ ⇒ $2\beta + \alpha = -1 + \frac{1}{2} = -\frac{1}{2}$ ⇒ $4\beta + 2\alpha = -1$ ...(ii) ∵ Point (α,β) lies on straight line $y=2x+11$ ∴ $\beta = 2\alpha + 11$ ⇒ $\beta - 2\alpha = 11$..(iii) On solving Eqs. (ii) and (iii), we get $5\beta = 10 \Rightarrow \beta = 2$ and $2\alpha = 2 - 11 = -9 \Rightarrow \alpha = -\frac{9}{2}$ ∴ Required point is $\left(-\frac{9}{2},2\right)$