Find the equation of the tangent to the circle $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$
which is parallel to $3 x - 4 y - 1 = 0$

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Solution

The equation of circle is $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$
Let the point of contact of circle be $(h, k)$
$⟹$ equation of tangent will be $h x + k y - (x + h) - 2 (y + k) - 4 = 0$
$⟹ (h - 1) x + (k - 2) y + (- h - 2 k - 4) = 0$
This equation is parallal to $3 x - 4 y - 1 = 0$
$h - 1 = 3 m ⟹ h = 3 m + 1$ and $k - 2 = - 4 m ⟹ k = 2 - 4 m$
$(h, k)$ lies on the circle $⟹ (3 m + 1)^{2} + (2 - 4 m)^{2} - 2 (3 m + 1) - 4 (2 - 4 m) - 4 = 0 ⟹ m = \pm \frac{3}{5}$
$⟹$ equation of tangent is $3 x - 4 y - 10 = 0$ and $3 x - 4 y + 20 = 0$

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(8, 1)

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