JEE PYQ: Circle Question 30
Question 30 - 2019 (10 Apr Shift 2)
If the line $ax + y = c$, touches both the curves $x^2 + y^2 = 1$ and $y^2 = 4\sqrt{2},x$, then $|c|$ is equal to
(1) $2$
(2) $\frac{1}{\sqrt{2}}$
(3) $\frac{1}{2}$
(4) $\sqrt{2}$
Show Answer
Answer: (4)
Solution
Tangent to $y^2 = 4\sqrt{2}x$: $y = mx + \frac{\sqrt{2}}{m}$, i.e., $mx - y + \frac{\sqrt{2}}{m} = 0$. Also tangent to $x^2 + y^2 = 1$: $\frac{\sqrt{2}/m}{\sqrt{m^2+1}} = 1 \Rightarrow 2 = m^2(m^2+1)$. So $m^2 = 1$, $c = \pm\sqrt{2}$, $|c| = \sqrt{2}$.
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