Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(I)}& \text{The length of the common chord of}& & \\ & \text{two circle of radii}3\text{and}4\text{units which}& & \\ & \text{intersect orthogonally is}\frac{k}{5},\text{then k is}& \text{(P)}& 1\\ \text{(II)}& \text{The circumference of the circle}& & \\ & x^2+y^2+4x+12y+p=0\text{is bisected}& & \\ & \text{by the circle}{x}^2+y^2-2x+8y-q=0,& \text{(Q)}& 2\\ & \text{then}p+q\text{equals to}& & \\ \text{(III)}& \text{The number of distinct chords of}& & \\ & x^2+y^2+2x+4y=0\text{which passes}& \left(R\right)& 12\\ & \text{through the point}(2,1)\text{and bisected}& & \\ & \text{by}x\text{- axis is}& & \\ \text{(IV)}& \text{One of the diameter of the circle}& \left(S\right)& 24\\ & \text{circumscribing the rectangle}ABCD& & \\ & \text{is}4y=x+7.\text{If}A\text{and}B\text{are}(-3,4)& & \\ & \text{and}(5,4)\text{respectively, then the area}& & \\ & \text{of the rectangle}ABCD\text{is}& & \\ & & \left(T\right)& 32\\ & & \left(U\right)& 36\end{array}$

Which of the following option is INCORRECT ?

• A. $\left(I\right)\to \left(R\right)$
• B. $\left(II\right)\to \left(U\right)$
• C. $\left(III\right)\to \left(P\right)$
• D. $\left(IV\right)\to \left(T\right)$

Answer 1

The correct option is A $\left(I\right)\to \left(R\right)$

$\left(I\right)$

$C_1{C}_2=\sqrt{3^2+4^2}=5$
Area of $△C_{1}P{C}_2$ is,
$\frac{1}{2}\times 4\times 3=\frac{1}{2}\times C_1{C}_2\times \frac{l}{2}\Rightarrow l=\frac{24}{5}\Rightarrow k=24$
$\left(I\right)\to \left(S\right)$

$\left(II\right)$
$S_1:x^2+y^2+4x+12y+p=0S_2:x^2+y^2-2x+8y-q=0$
Equation of the common chord is
$S_1-S_2=0$
$\Rightarrow 6x+4y+p+q=0$
Common chord will pass through the center of $S_1$,
$C_1=(-2,-6)\Rightarrow -12-24+p+q=0\therefore p+q=36$
$\left(II\right)\to \left(U\right)$

$\left(III\right)$
Equation of the circle is
$x^2+y^2+2x+4y=0$
Let $(a,0)$ be the mid point of the chord, then equation of the chord,
$T=S_{1}ax+(x+a)+2y=a^2+2a$
It passes through the point $(2,1)$
$\therefore 2a+2+a+2=a^2+2a\Rightarrow a^2-a-4=0\Rightarrow a=\frac{1\pm \sqrt{17}}{2}$
Since, $(\frac{1+\sqrt{17}}{2},0)$ lies outside the circle.
$\therefore$ Number of chord is $1.$
$\left(III\right)\to \left(P\right)$

$\left(IV\right)$
$A=(-3,4)$ and
$B=(5,4)$


Midpoint of $AB=(1,4)$
Line $AB$ is parallel to $x-$ axis
So, the equation of the perpendicular bisector of $AB$ is
$x=1$
Diameter is
$4y=x+7$
Therefore, the center of the circle $\left(C_1\right)$ is
$4y=1+7\Rightarrow y=2C_1=(1,2)$
Radius $=\sqrt{4^2+2^2}=\sqrt{20}$
Sides of the rectangle are,
$AB=8BC=2\sqrt{20-4^2}=4$
$\therefore Area=4\times 8=32$
$\left(IV\right)\to \left(T\right)$

Hence,
$\left(I\right)\to \left(S\right)$
$\left(II\right)\to \left(U\right)$
$\left(III\right)\to \left(P\right)$
$\left(IV\right)\to \left(T\right)$