Given the circle $C$ with the equation $x^2+y^2-2 x+10 y-38=0$. Match the List I with the List II given…
Given the circle $C$ with the equation $x^2+y^2-2 x+10 y-38=0$. Match the List I with the List II given below concerning $C$
The correct answer is
$\begin{array}{llll}\text { A } & \text { B } & \text { C } & \text { D }\end{array}$
$\begin{array}{llll}\text { III } & \text { I } & \text { V } & \text { II }\end{array}$
$\begin{array}{llll}\text { IV } & \text { V } & \text { I } & \text { II }\end{array}$
$\begin{array}{llll}\text { III } & \text { V } & \text { I } & \text { II }\end{array}$
$\begin{array}{llll}\text { IV } & \text { II } & \text { I } & \text { V }\end{array}$
Solution
Given equation of circle is $x^2+y^2-2 x+10 y-38=0$
(A) Polar equation at point $(4,3)$ is $S_1=0$.
$\begin{aligned} & \Rightarrow \quad x \times 4+y \times 3-(x+4) \\ & \quad+5(y+3)-38=0 \\ & \Rightarrow \quad 3 x+8 y=27\end{aligned}$
(B) Equation of tangent at point $(9,-5)$ is
$\begin{array}{crrrl}x \times 9+y \times(-5)-(x+9)+5(y-5)-38 & =0 \\ \Rightarrow & & 8 x-72 & =0 \\ \Rightarrow & & x & =9\end{array}$
(C) On differentiating given equation w.r.t. $x$, we get
$2 x+2 y \frac{d y}{d x}-2+10 \frac{d y}{d x}-0=0$
$\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}(2 y+10)=2-2 x \\ & \Rightarrow\left(\frac{d y}{d x}\right)_{(-7,-5)}=\frac{2-2 \times(-7)}{2 \times(-5)+10}=\frac{16}{0}=\frac{1}{0}\end{aligned}$
$\therefore$ Equation of normal at point $(-7,-5)$ is
$\begin{aligned} & y+5=-0(x+7) \\ & \Rightarrow \quad y+5=0 \\ & \end{aligned}$
(D) Centre of circle $C$ is $(1,-5)$
$\therefore$ Equation of diameter passing through $(1,-5)$ and $(1,3)$ is
$\begin{array}{rlrl} & y+5 & =\frac{3+5}{1-1}(x-1) \\ \Rightarrow & & y+5 & =\frac{8}{0}(x-1) \\ \Rightarrow & x-1 & =0 \\ \Rightarrow & x & =1\end{array}$
