Question

Given the circle $C$ with the equation $x^2+y^2-2x+10y-38=0.$ 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
$x^2+y^2-2x+10y-38=0$
(A) Polar equation at point $(4,3)$ is $S_1=0$.
$\Rightarrow x\times 4+y\times 3-(x+4)$
$+5(y+3)-38=0$
$\Rightarrow 3x+8y=27$
(B) Equation of tangent at point $(9,-5)$ is
$x\times 9+y\times (-5)-(x+9)+5(y-5)-38=0$
$\Rightarrow 8x-72=0$
$\Rightarrow x=9$
(C) On differentiating given equation w.r.t. $x$, we get
$2x+2y\frac{dy}{dx}-2+10\frac{dy}{dx}-0=0$
$\Rightarrow \frac{dy}{dx}(2y+10)=2-2x$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(-7,-5)}=\frac{2-2\times (-7)}{2\times (-5)+10}=\frac{16}{0}=\frac{1}{0}$
$\therefore$ 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
$y+5=\frac{3+5}{1-1}(x-1)$
$\Rightarrow y+5=\frac{8}{0}(x-1)$
$\Rightarrow x-1=0$
$\Rightarrow x=1$