Q

# Solve this problem - Limit , continuity and differentiability - JEE Main-23

The angle of intersection between $x^2 + y^2 = 1$ and $x + y =1$ is $\theta$ at a point, where abscissa is positive, then $\tan\theta$ equals ?

• Option 1)

1

• Option 2)

2

• Option 3)

3

• Option 4)

4

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As we have learned

Angle of intersection of two curves -

The angle of intersection of two curves is the angle subtended between the tangents at their point of intersection.Let  m1  &  m2 are two slope of tangents at intersection point of two curves then

$tan\theta=\frac{[m_{1}-m_{2}]}{1+m_{1}m_{2}}$

- wherein

where $\theta$ is angle between two curves tangents.

$x^{2}+y^{2}=1$   and  x+y =1

$\Rightarrow x^{2}+(1-x)^{2}=1 \Rightarrow 2x^{2}-2x=0 \Rightarrow 0=0,1$

$\therefore$  point of intersection are (0,1) and (1,0)  we need to find angle at (1,0)

$x^{2}+y^{2}=1 \Rightarrow 2x +2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-x/y$

$\Rightarrow \frac{dy}{dx}$ at ( 1,0) will be infinite , i.e tangent will be vertical

$x+y=1\Rightarrow \frac{dy}{dx} =-1$

$\Rightarrow x+y =1$   makes $135 \degree$ with x-axis so with a vertical line it will  make $45 \degree$

$\therefore \tan \theta = \tan 45 \degree =1$

Option 1)

1

Option 2)

2

Option 3)

3

Option 4)

4

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