Question

$\begin{array}{cc}Column1& Column2\\ \text{a. If}ax+by-5=0\text{is the equation of the}\text{chord of the circle}& \text{p.}6\\ (x-3{)}^2+(y-4{)}^2=4,\text{which passes through}(2,3)\text{and at}& \\ \text{the greatest distance from the centre, then}|a+b|\text{is equal to}& \\ \text{b. Let}O\text{be the origin and}P\text{be a variable}\text{point on the circle}& \text{q.}3\\ x^2+y^2+2x+2y=0.\text{If the locus of midpoint of}OP\text{is}& \\ x^2+y^2+2gx+2fy+c=0,\text{then}(g+f)\text{is equal to}& \\ \text{c. The x–coordinate of the centre of the}\text{smallest circle which}& \text{r.}2\\ \text{cuts the circles}{x}^2+y^2-2x-4y-4=0\text{and}{x}^2+y^2-10x& \\ +12y+52=0\text{orthogonally is}& \\ \text{d. If}\theta \text{be the angle between two tangents}\text{which are drawn}& \text{s.}1\\ \text{to the circles}{x}^2+y^2-6\sqrt{3}x-6y+27=0\text{from the origin,}& \\ \text{then}2\sqrt{3}\tan \theta \text{equals to}& \end{array}$

Which of the following is correct?

• A. $a\to s,b\to r,c\to p,d\to q$
• B. $a\to r,b\to s,c\to p,d\to q$
• C. $a\to s,b\to r,c\to q,d\to p$
• D. $a\to r,b\to s,c\to q,d\to p$

Answer 1

The correct option is D $a\to r,b\to s,c\to q,d\to p$

$a.$
We know that $(2,3)$ lies inside the circle,
The chord which is at the greatest distance from the centre is bisected at $(2,3),$

Slope of line joining point $(2,3)$ and centre is
$m=\frac{4-3}{3-2}=1$
The chord will be perpendicular to this, so th slope of the chord
$=-1$
Equation of chord is
$y-3=-(x-2)\Rightarrow x+y-5=0$
Comparing with $ax+by-5=0$, we get
$\frac{a}{1}=\frac{b}{1}=\frac{-5}{-5}\Rightarrow a=b=1\therefore |a+b|=2$

$b.$
Let P be the point $(\alpha ,\beta )$
Assuming the midpoint be $M(h,k)$, so
$(h,k)=(\frac{\alpha }{2},\frac{\beta }{2})\Rightarrow \alpha =2h,\beta =2k$
Then $P=(2h,2k)$
So putting the point $P$ in the equation of circle, we get
$4h^2+4k^2+4h+4k=0\Rightarrow h^2+k^2+h+k=0$
Therefore, the locus of the midpoint is
$x^2+y^2+x+y=0$
Comparing with $x^2+y^2+2gx+2fy+c=0$, we get
$f=g=\frac{1}{2},c=0\therefore f+g=1$

$c.$
Given circles are
$x^2+y^2-2x-4y-4=0$
$C_1=(1,2),r_1=3$
$x^2+y^2-10x+12y+52=0$
$C_2=(5,-6),r_2=3$

The equation of line joinig $C_1{C}_2$ is
$y-2=\frac{-6-2}{5-1}(x-1)\Rightarrow y-2=-2x+2\Rightarrow 2x+y-4=0\cdots \left(1\right)$
The equation of the radical axis is
$8x-16y-56=0\Rightarrow x-2y-7=0\cdots \left(2\right)$
Finding the point of intersection of $\left(1\right)$ and $\left(2\right)$, we get
$5x-15=0\Rightarrow x=3$
Hence, the required $x-$ coordinate is $3$.

$d.$
Given circle is
$x^2+y^2-6\sqrt{3}x-6y+27=0$
$C=(3\sqrt{3},3),r=3$
Distance between centre and origin
$d=\sqrt{27+9}=6$
Now, we know that
$\sin \frac{\theta }{2}=\frac{r}{d}\Rightarrow \sin \frac{\theta }{2}=\frac{3}{6}=\frac{1}{2}\Rightarrow \frac{\theta }{2}=\frac{\pi }{6}\Rightarrow \theta =\frac{\pi }{3}\Rightarrow \tan \theta =\sqrt{3}\therefore 2\sqrt{3}\tan \theta =6$