Question

The angle between the tangents to the circles x^2+ y^2 = 1 and x^2 + (y - 1)^2 = 2 at the point (1; 0) is

Answer 1

Consider the circle of equation,

$\displaystyle\sf{\longrightarrow x^2+y^2=1}$

Differentiating wrt $\displaystyle\sf {x,}$

$\displaystyle\sf{\longrightarrow 2x+2y\cdot y'=0}$

$\displaystyle\sf{\longrightarrow y'=-\dfrac {x}{y}}$

This is the slope of tangent to this circle.

$\displaystyle\sf{\longrightarrow m_1=-\dfrac {x}{y}}$

At $\displaystyle\sf {(x,\ y)=(1,\ 0),}$

$\displaystyle\sf{\longrightarrow m_1=-\dfrac {1}{0}}$

This implies tangent to this circle at (1, 0) is parallel to y axis since denominator of slope is zero.

That means this tangent makes 90° with x axis.

$\displaystyle\sf{\longrightarrow \theta_1=90^o}$

Consider the circle of equation,

$\displaystyle\sf{\longrightarrow x^2+(y-1)^2=2}$

Differentiating wrt $\displaystyle\sf {x,}$

$\displaystyle\sf{\longrightarrow 2x+2(y-1)\cdot y'=0}$

$\displaystyle\sf{\longrightarrow y'=\dfrac {x}{1-y}}$

This is slope of tangent to this circle.

$\displaystyle\sf{\longrightarrow m_2=\dfrac {x}{1-y}}$

At $\displaystyle\sf {(x,\ y)=(1,\ 0),}$

$\displaystyle\sf{\longrightarrow m_2=\dfrac {1}{1-0}}$

$\displaystyle\sf{\longrightarrow m_2=1}$

Angle made by this tangent with x axis,

$\displaystyle\sf{\longrightarrow \theta_2=\tan^{-1}(1)}$

$\displaystyle\sf{\longrightarrow\theta_2=45^o}$

Then, angle between the tangents,

$\displaystyle\sf{\longrightarrow \theta=90^o-45^o}$

$\displaystyle\sf {\longrightarrow\underline {\underline {\theta=45^o}}}$