Using integration find the area of the triangle formed by positive x-axis and tangent and normal of the circle
`x^2+y^2=4 at (1, sqrt3)`
The given equation of the circle is x2+y2=4.
The equation of the normal to the circle at (1,√3) is same as the line joining the points (1,√3) and (0, 0), which is given by
So, the slope of normal is `sqrt3.`
We know that the product of the slopes of the normal and the tangent is −1
Therefore, the slope of tangent is `−1/sqrt3`
Now, the equation of the tangent to the circle at (1,√3) is given by
Putting y = 0 in (2), we get x = 4.
Thus, ABC is the triangle formed by the positive x-axis and tangent and normal to the given circle at `(1,sqrt3)`
Area of ∆AOB = Area of ∆AOM + Area of ∆AMB
Thus, the area of the triangle so formed is `2sqrt3` square units.