# Find the equation of normal and tangent to the circle x^2+y^2=4 at the point (2cos45^@,2sin45^@)?

Apr 11, 2017

Equation of normal is $y = x$ and that of tangent is $x + y = 2 \sqrt{2}$

#### Explanation:

We are seeking a tangent and normal from a point $\left(2 \cos {45}^{\circ} , 2 \sin {45}^{\circ}\right)$ i.e. $\left(\frac{2}{\sqrt{2}} , \frac{2}{\sqrt{2}}\right)$ or $\left(\sqrt{2} , \sqrt{2}\right)$.

Normal to a point on a circle is the line joining center to the given point. Asthe center of ${x}^{2} + {y}^{2} = 4$ is $\left(0 , 0\right)$ and te point is $\left(\sqrt{2} , \sqrt{2}\right)$, the equation of normal is

$\frac{y - 0}{\sqrt{2} - 0} = \frac{x - 0}{\sqrt{2} - 0}$ or $\frac{y}{\sqrt{2}} = \frac{x}{\sqrt{2}}$ or $y = x$.

Its slope is $1$. As normal and tangent are perpendicular to each other, product of their slopes is $- 1$ and hence slope of the tangent is $\frac{- 1}{1} = - 1$

So our tangent passes through $\left(\sqrt{2} , \sqrt{2}\right)$ and has a slope of $- 1$. Therefore its equation is

$y - \sqrt{2} = - 1 \left(x - \sqrt{2}\right)$ or $x + y = 2 \sqrt{2}$

graph{(x+y-2sqrt2)(x-y)(x^2+y^2-4)=0 [-4.667, 5.333, -2.32, 2.68]}