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Equation of Tangent in Slope Form

If a line y =...

Question

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:

c^{2} + 7 c + 6 = 0

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c^{2} - 6 c + 7 = 0

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c^{2} - 7 c + 6 = 0

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c^{2} + 6 c + 7 = 0

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Solution

The correct option is D $c^{2} + 6 c + 7 = 0$
The equation of tangent $(L_{1})$ to $x^{2} + y^{2} = 1$ at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ is
$x x_{1} + y y_{1} - 1 = 0 \Rightarrow \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} - 1 = 0 \Rightarrow x + y - \sqrt{2} = 0$
Slope of $L_{1}$ is
$m_{L_{1}} = - 1$
Slope of perpendicular line is
$m = 1$

Given line is $y = m x + c$ , so
$y = x + c$ is tangent to $(x - 3)^{2} + y^{2} = 1$
Centre and radius are
$C = (3, 0), r = 1$
We know that the perpendicular distance of tangent from centre is equal to radius, so
$∣ ∣ ∣ \frac{3 + c}{\sqrt{2}} ∣ ∣ ∣ = 1 \Rightarrow c + 3 = \pm \sqrt{2}$
Squaring both sides, we get
$\Rightarrow c^{2} + 6 c + 9 = 2 ∴ c^{2} + 6 c + 7 = 0$

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