#### please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1 sub question 4 maths textbook solution

Answer: $\frac{\pi}{4}$

Hint – The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

m1=slope of first curve.   m2=slope of second curve.

Given – $x^{2}+y^{2}-4 x-1=0 \ldots \text { (1) }$  &

$x^{2}+y^{2}-2 y-9=0 \ldots(2)$

First curve is  $x^{2}+y^{2}-4 x-1=0$
\begin{aligned} &x^{2}-4 x+4+y^{2}-4-1=0 \\ &(x-2)^{2}+y^{2}-5=0 \ldots \ldots \text { (3) } \end{aligned}

Subtracting (2) from (1) we get ,

\begin{aligned} &=x^{2}+y^{2}-4 x-1-\left(x^{2}+y^{2}-2 y-9\right)=0 \\ &=x^{2}+y^{2}-4 x-1-x^{2}-y^{2}+2 y+9=0 \\ &=-4 x-1+2 y+9=0 \\ &=-4 x+2 y+8=0 \\ &=2 y=4 x-8 \\ &=y=2 x-4 \end{aligned}

Substituting y=2x-4 in (3), we get
\begin{aligned} &=(x-2)^{2}+(2 x-4)^{2}-5=0 \\ &=(x-2)^{2}+4(x-2)^{2}-5=0 \\ &=(x-2)^{2}+(1+4)-5=0 \\ &=5(x-2)^{2}-5=0 \end{aligned}

\begin{aligned} &=5\left((x-2)^{2}-1\right)=0 \\ &=(x-2)^{2}-1=0 \\ &=(x-2)^{2}=1 \\ &=(x-2)=\pm 1 \end{aligned}

\begin{aligned} &=x=1+2 \text { or } x=-1+2 \\ &=x=3 \text { or } x=1 \end{aligned}

so When x =3
\begin{aligned} &y=2 \times 3-4 \\ &y=6-4 \\ &y=2 \end{aligned}

when x=1

\begin{aligned} &y=2 \times 1-4 \\ &y=2-4 \\ &y=-2 \end{aligned}

The point of intersection of two curves are (3, 2) & (1,-2).

Differentiating curves (1) & (2) with respect to x,

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0 \\ &=x^{2}+y^{2}-4 x-1=0 \\ &=2 x+\frac{2 y(d y)}{d x}-4-0=0 \end{aligned}

\begin{aligned} &=x+y \frac{d y}{d x}-2=0 \\ &=y \frac{d y}{d x}=2-x \\ &m_{1}=\frac{d y}{d x}=\frac{2-x}{y} \ldots \ldots \text { (4) } \end{aligned}

Second curve is $x^{2}+y^{2}-2 y-9=0$
\begin{aligned} &2 x+\frac{2 y(d y)}{d x}-\frac{2 d y}{d x}-0=0 \\ &=x+y \frac{d y}{d x}-\frac{d y}{d x}=0 \\ &=x+(y-1) \frac{d y}{d x}=0 \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{y-1} \ldots \ldots(5) \end{aligned}

At (3, 2) in eq (4), we get

$m_{1}=\frac{d y}{d x}=\frac{2-3}{2}=\frac{-1}{2}$

At (3, 2) in eq (5), we get

$m_{2}=\frac{d y}{d x}=\frac{-3}{2-1}=-3$

At (1,-2) in eq (4), we get

\begin{aligned} &m_{1}=\frac{d y}{d x}=\frac{2-(1)}{-2}=\frac{2}{-2}=-1 \\ &m_{2}=\frac{d y}{d x}=\frac{-x}{y-1}=\frac{-1}{-2-1}=\frac{-1}{-3}=\frac{1}{3} \end{aligned}

Value of  $m_{1}=\frac{-1}{2} \&-1$

Value of $m_{2}=-3 \& \frac{1}{3}$

Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

When m1is$\frac{-1}{2}$ and m2 is-3

\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{-\frac{1}{2}-(-1)}{1+\left(-\frac{1}{2}\right)(-1)}\right| \\ &\operatorname{tan} \theta=|-1| \\ &\operatorname{tan} \theta=1 \quad:: \tan \left(\frac{\pi}{4}\right)=1 \\ &=\frac{\pi}{4} \end{aligned}

When m1=-1 and m2=$\frac{1}{3}$
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{-1-\left(\frac{1}{3}\right)}{1+(-1)\left(\frac{1}{3}\right)}\right|=\left|\frac{-1+\frac{1}{3}}{1-\frac{1}{3}}\right| \\ &\operatorname{tan} \theta=|-1| \\ &\operatorname{tan} \theta=1 \\ &=\theta=\frac{\pi}{4} \end{aligned}