JEE PYQ: Circle Question 21
Question 21 - 2020 (08 Jan Shift 2)
If a line, $y = mx + c$ is a tangent to the circle, $(x-3)^2 + y^2 = 1$ and it is perpendicular to a line $L_1$, where $L_1$ is the tangent to the circle, $x^2 + y^2 = 1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$; then:
(1) $c^2 - 7c + 6 = 0$
(2) $c^2 + 7c + 6 = 0$
(3) $c^2 + 6c + 7 = 0$
(4) $c^2 - 6c + 7 = 0$
Show Answer
Answer: (3)
Solution
$L_1$: tangent to $x^2+y^2=1$ at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ is $x + y = \sqrt{2}$, slope $= -1$. Perpendicular: $m = 1$. Tangent $y = x + c$ to $(x-3)^2 + y^2 = 1$: $\frac{|c+3|}{\sqrt{2}} = 1 \Rightarrow (c+3)^2 = 2 \Rightarrow c^2 + 6c + 7 = 0$.
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