Q42 / 75 · JEE Main 2020 (08 Jan Shift 2) · Mathematics

If a line $y = m x + c$ , 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 $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ , then

A
$c^{2} - 7 c + 6 = 0$
B
$c^{2} + 7 c + 6 = 0$
Correct answer
C
$c^{2} + 6 c + 7 = 0$
D
$c^{2} - 6 c + 7 = 0$

Solution

Slope of tangent to $x^{2} + y^{2} = 1$ at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
$x^{2} + y^{2} = 1$
$2 x + 2 y y^{'} = 0$
$y^{'} = - \frac{x}{y} = - 1$
Given that, $y = m x + c$ , is perpendicular to $L_{1}$ ,
So, $m = 1 \Rightarrow y = x + c$

Now distance of $(3, 0)$ from $y = x + c$ is
$|\frac{c + 3}{\sqrt{2}}| = 1$
$c^{2} + 6 c + 9 = 2$
$c^{2} + 6 c + 7 = 0$ .

Hence, the correct answer is option: 'C'.

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If a line y=mx+c, is a tangent to the circle x-32+y2=1, and it is perpendicular to a line L1, where L1 is the tangent to the circle x2+y2=1, at the point 12, 12, then

Solution

Relevant Questions

If a line $y = m x + c$ , 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 $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ , then