Question

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

Answer 1

Equation of normal is `y=x` and that of tangent is `x+y=2sqrt2`

Explanation:

We are seeking a tangent and normal from a point `(2cos45^@,2sin45^@)` i.e. `(2/sqrt2,2/sqrt2)` or `(sqrt2,sqrt2)`.

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 `(0,0)` and te point is `(sqrt2,sqrt2)`, the equation of normal is

`(y-0)/(sqrt2-0)=(x-0)/(sqrt2-0)` or `y/sqrt2=x/sqrt2` 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 `(-1)/1=-1`

So our tangent passes through `(sqrt2,sqrt2)` and has a slope of `-1`. Therefore its equation is

`y-sqrt2=-1(x-sqrt2)` or `x+y=2sqrt2`

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