Question

Match the following by approximately matching the lists based on the information given in Column I and Column II

$\begin{array}{cc}Column1& Column2\\ \text{a. The length of the common chord of two circles of radii}3\text{and}& \text{p.}1\\ 4\text{units which intersect orthogonally is}\frac{k}{5},\text{then}k\text{is equal to}& \\ \text{b. The circumference of the circle}{x}^2+y^2+4x+12y+p=0& \text{q.}24\\ \text{is bisected by the circle}{x}^2+y^2-2x+8y-q=0\text{, then}& \\ p+q\text{is equal to}& \\ \text{c. Number of distinct chords of the circle}2x(x-\sqrt{2})+y(2y-1)& \text{r.}32\\ =0\text{chords are passing through the point}(\sqrt{2},\frac{1}{2})\text{and are}& \\ \text{bisected on x-axis is}& \\ \text{d. One of the diameters of the circle circumscribing the rectangle}& \text{s.}36\\ ABCD\text{is}4y=x+7\text{. If}A\text{and}B\text{are the points}(-3,4)\text{and}& \\ (5,4)\text{respectively,}\text{then the area of rectangle is}& \end{array}$

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

Answer 1

The correct option is A $a-q,b-s,c-p,d-r$

$\left(a\right)$
Let the length of the common chord be $2a$. Then, $\sqrt{9-a^2}+\sqrt{16-a^2}=5$
or $\sqrt{16-a^2}=5-\sqrt{9-a^2}$
or $16-a^2=25+9-a^2-10\sqrt{9-a^2}$
or $10\sqrt{9-a^2}=18$
or $100(9-a^2)=324$
or $100a^2=576$
or $a=\sqrt{\frac{576}{100}}=\frac{24}{10}$
$\therefore 2a=\frac{24}{5}=\frac{k}{5}\Rightarrow k=24$

$\left(b\right)$
The equation of common chord is $6x+4y+p+q=0$
The common chord passes through the center $(-2,-6)$ of the circle $x^2+y^2+4x+12y+p=0$. Therefore, $p+q=36$

$\left(c\right)$
The equation of the circle is $2x^2+2y^2-2\sqrt{2}x-y=0$
let $(\alpha ,0)$ be the midpoint of a chord. Then, the equation of the chord is $2\alpha x-\sqrt{2}(x+\alpha )-\frac{1}{2}(y+0)=2{\alpha }^2-2\sqrt{2}\alpha$
Since it passes through the point $(\sqrt{2},\frac{1}{2})$, we have $2\sqrt{2}\alpha -\sqrt{2}(\sqrt{2}+\alpha )-\frac{1}{4}=2{\alpha }^2-2\sqrt{2}\alpha$
$\Rightarrow 8{\alpha }^2-12\sqrt{2}\alpha +9=0$
$\Rightarrow (2\sqrt{2}\alpha -3{)}^2=0$
$\alpha =\frac{3}{2\sqrt{2}},\frac{3}{2\sqrt{2}}$
Therefore, the number of chords is $1$

$\left(d\right)$
Midpoint of $AB\equiv (1,4)$
The equation perpenduicular bisector of $AB$ is $x=1$
A dimeter of the circle is $4y=x+7$
Therefore, the center of the circle is $(1,2)$
Hence, the sides of the rectangle are $8$ and $4$. Therefore,
Area $=32$