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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} + 6 c + 7 = 0

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

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Solution

The correct option is C $c^{2} + 6 c + 7 = 0$

Equation of $L_{1} : \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1$
$\Rightarrow x + y = \sqrt{2}$
Now $L ⊥ L_{1}$
$\Rightarrow m \times m_{1} = - 1$
$\Rightarrow m \times - 1 = - 1$
$\Rightarrow m = 1$
$∴$ Slope of $L = 1$
$\Rightarrow$ Equation of $L : y = x + c$
Now as $L$ is tangent to $S : (x - 3)^{2} + y^{2} = 1,$
$⊥^{r}$ distance from $O$ to $L =$ radius
$\Rightarrow ∣ ∣ ∣ \frac{c + 3}{\sqrt{2}} ∣ ∣ ∣ = 1$
$\Rightarrow | c + 3 | = \sqrt{2}$
$\Rightarrow c^{2} + 9 + 6 c = 2$
$\Rightarrow c^{2} + 6 c + 7 = 0$

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Equation of Tangent in Slope Form

Standard XII Mathematics

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If a line y=mx+c is a tangent to the circle (x−3)2+y2=1 and it is perpendicular to a line L1, where L1 is the tangent to the circle x2+y2=1 at the point (1√2,1√2), then

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