# Using integration find the area of the triangle formed by positive x-axis and tangent and normal to the circle x2 + y2 = 4 at (1, √3).OREvaluate: as a limit of a sum.  [CBSE 2015]

As first we need to trace the area to be determined. As x2+y2 = 4 represents a circle whose centre is at (0,0) and radius = 2 cm. The rough sketch is shown below:

As normal at (1,√3) passes through origin too because in a circle a normal always passes through centre of the circle.

Equation of normal is y = √3 x

Similarly equation of tangent can be written using one point and slope form.

As, x2 + y2 = 4

Differentiating w.r.t x:     We need to determine the area enclosed i.e. area(region ABC).

Area(region ABC) = area(region ABD) + area(region BDC)

Area = area under curve y = √3x + area under Required area = Area = Area = Area = Required area = OR

We know that a definite integral can be evaluated as a limit of a sum as- Where h = As we have to find: Let I = and on comparing I with the formula we can say that a = 1 and b = 3.

I = I = 2 I = 2 I = 2 Each bracket contains ‘n’ terms.

I = 2 I = 2 I = 2 Using formula for sum of first n natural numbers; sum of squares of first n natural numbers and sun of n terms of a GP we get:

I = 2 I = 2 As h = I = 2 I = 2 Using algebra of limits:

I = 2 Use the formula: I = 2 I = Rate this question :

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