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:

List - I	List - II
(i) The equation of the polar of $(4,3)$ with respect to C	(a) $y+5=0$
(ii) The equation of the tangent at $(9,-5)$ on C	(b) $x=1$
(iii) The equation of the normal at $(-7,-5)$ on C	(c) $3x+8y=27$
(iv) The equation of the diameter of C passing through $(1,3)$	(d) $x+y=3$
	(e) $x=9$

• A. $\left(i\right)-\left(c\right),\left(ii\right)-\left(a\right),\left(iii\right)-\left(e\right),\left(iv\right)-\left(b\right)$
• B. $\left(i\right)-\left(d\right),\left(ii\right)-\left(e\right),\left(iii\right)-\left(a\right),\left(iv\right)-\left(b\right)$
• C. $\left(i\right)-\left(c\right),\left(ii\right)-\left(e\right),\left(iii\right)-\left(a\right),\left(iv\right)-\left(b\right)$
• D. $\left(i\right)-\left(d\right),\left(ii\right)-\left(b\right),\left(iii\right)-\left(a\right),\left(iv\right)-\left(c\right)$

Answer 1

The correct option is C $\left(i\right)-\left(c\right),\left(ii\right)-\left(e\right),\left(iii\right)-\left(a\right),\left(iv\right)-\left(b\right)$

Given eqn is $s:(x-1{)}^2+(y-5{)}^2=8^2$

i) polar of (4, 3) is $s_1=0$

$\Rightarrow (x-1)(4-1)+(y+5)(3+5)=8^2$

$\Rightarrow 3x+8y=27$

ii) Tangent at (9, -5) is $s_1=0$

$\Rightarrow x-9=0$

iii) Equation of normal at (-7, -5) as centre is $(1,-5)$

$\frac{x+7}{-7+(-1)}=\frac{y+5}{-5+5}$

$\Rightarrow y+5=0$

iv) centre is (1, -5). so eqn of diameter is

$\Rightarrow \frac{(y-3)}{x-1}=\frac{-5-3}{1-1}$

$\Rightarrow x-1=0$