Find an equation of the tangent line to the circle x2+y2=9 at the point (2, √5)

Find an equation of the tangent line to the circle x2+y2=9 at the point (2, √5)

The radius r of the circle is 3, because $$x^2+y^2=r^2$$  and $$r^2 = 9$$

The distance of the point of the co-ordinate origin amounts    $$\sqrt{2^2+\sqrt{5}^2}=\sqrt{4+5}=\sqrt{9}=3 = r$$

so the point lies on the circle: The equation is: $$y=-\dfrac{x_p}{y_p}\left( x-x_p \right) +y_p$$  or $$y=-\dfrac{x_p}{y_p}x +\dfrac{r^2}{y_p}$$

so   $$x_p = 2 \quad y_p=\sqrt{5}\\ y=-\dfrac{2}{\sqrt{5}}x +\dfrac{r^2}{\sqrt{5}}$$ finally   $$y=-\dfrac{2}{\sqrt{5}}x +\dfrac{9}{\sqrt{5}}$$

Rewrite as y = (9 - x²)^1/2

The slope at any point x is then given by slope = -x/(9 - x²)^1/2

When x = 2 the slope is -2/(9 - 2²)^1/2 = -2/√5

The equation of a straight line is y = m*x + c where m is the slope and c is a constant (the y-intercept).  We know the slope and can find c because we know that when x = 2, y = √5, so

√5 = (-2/√5)*2 + c

c = √5+4/√5 or c = √5*(1+4/5) 0r c = 1.8*√5

So  y = (-2/√5)*x + 1.8*√5

or y = √5*(-2/5)x + 1.8*√5

or y = -0.4*√5*x + 1.8*√5