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Equation of Tangent in Slope Form

If the tangen...

Question

If the tangent drawn at $(1, 2)$ to the circle $x^{2} + y^{2} - 4 x + 2 y - 5 = 0$ is normal to the circle $x^{2} + y^{2} - 4 x + k y - 1 = 0$ , then the value of $| 3 k |$ is

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Solution

Given circle is $x^{2} + y^{2} - 4 x + 2 y - 5 = 0$
Equation of tangent at $(1, 2)$ is
$x x_{1} + y y_{1} - 2 (x + x_{1}) + (y + y_{1}) - 5 = 0 \Rightarrow x + 2 y - 2 (x + 1) + (y + 2) - 5 = 0 \Rightarrow x - 3 y + 5 = 0$

As it is normal to the circle $x^{2} + y^{2} - 4 x + k y - 1 = 0$ , so it passes through the centre of circle i.e. $(2, - \frac{k}{2})$
$\Rightarrow 2 + \frac{3 k}{2} + 5 = 0$
$\Rightarrow k = - \frac{14}{3}$
$∴ | 3 k | = 14$

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Equation of Tangent in Slope Form

Standard XII Mathematics

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