Question

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 $C$

	Lis I		Lis II
A	The equation of the polar of $(4, 3)$ with respect to C	i	$y +5 =O$
B	The equation of the tangent at $(9, − 5)$ on C	ii	$x=1$
C	The equation of the normal at $(− 7, − 5)$ on $C$	iii	$3x+8y=27$
D	The equation of the diameter of $C$ passing through $(1, 3)$	iv	$x + y = 3$
		v	$x= 9$

The correct answer is

• A. $A \to III, B \to I, C \to V, D \to II$
• B. $A \to IV, B \to V, C \to I, D \to II$
• C. $A \to III, B \to V, C \to I, D \to II$
• D. $A \to IV, B \to II, C \to I, D \to V$

Answer 1

Answer: (C)

Given equation of circle is
$x^{2}+y^{2}-2 x+10 y-38=0$
(A) Polar equation at point $(4,3)$ is $S_{1}=0$.
$\Rightarrow x \times 4+y \times 3-(x+4)$
$+5(y+3)-38=0$
$\Rightarrow 3 x+8 y=27$
(B) Equation of tangent at point $(9,-5)$ is
$x \times 9+y \times(-5)-(x+9)+5(y-5)-38=0$
$\Rightarrow 8 x-72=0$
$\Rightarrow x=9$
(C) On differentiating given equation w.r.t. $x$, we get
$2 x+2 y \frac{d y}{d x}-2+10 \frac{d y}{d x}-0=0$
$\Rightarrow \frac{d y}{d x}(2 y+10)=2-2 x$
$\Rightarrow \left(\frac{d y}{d x}\right)_{(-7,-5)}=\frac{2-2 \times(-7)}{2 \times(-5)+10}=\frac{16}{0}=\frac{1}{0}$
$\therefore $ Equation of normal at point $(-7,-5)$ is
$ y+5 =-0(x+7) $
$\Rightarrow y+5 =0$
(D) Centre of circle $C$ is $(1,-5)$.
$\therefore $ Equation of diameter passing through
$(1,-5)$ and $(1,3)$ is
$ y+5 =\frac{3+5}{1-1}(x-1) $
$\Rightarrow y+5 =\frac{8}{0}(x-1) $
$ \Rightarrow x-1 =0 $
$ \Rightarrow x =1$