Find the equations of the tangents to the circle $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ at the point (4,2).

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Solution

The equation of any straight line passing through point $(4, 2)$ is
$y - 2 = m (x - 4) ⟶ (1)$
From $(1)$ is target to the given circle provided the length of the perpendicular from center of given circle to $(1)$ is equal to radius of the circle
$x^{2} + y^{2} - 4 x - 4 y + 4 = 0$
$2 g = - 4 \Rightarrow g = - 2$
$2 f = - 4 f = - 2$
Center $= (- g, - f) = (2, 2)$
$r = \sqrt{g^{2} + f^{2} - c} = \sqrt{4 + 4 + 4} = 2 m = \frac{y - 2}{x - 4}$
Slope of tangent $x - 4 = 0$ $x = 4$ Equation of tangent (i) $x = 4$

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