Question

Column I		Column II
A	The length of the common chord of two circles of radii 3 and 4 units which intersect orthogonally is $\frac{k}{5}$, then $k$ equals to	p	1
B	The circumference of the circle $x^2+y^2+4x+12y+p=0$ is bisected by the circlc $x^2+y^2-2x+8y-q=0$, then $p+q$ is equal to	q	24
C	Number of distinct chords of the circles $2x(x-\sqrt{2})+y(2y-1)=0$ is passing through the point $\left(\sqrt{2},\frac{1}{2}\right)$ and are bisected by $x$-axis is	r	32
D	One of the diameters of the circles circumscribing the rectangle $ABCD$ is $4y=x+7$. If $A$ and $B$ are the points $(-3,4)$ and $(5,4)$ respectively, then the area of the rectangle is equal to	s	36

• A. (A) -(p), (B) - (q), (C) - (r), (D) - (s)
• B. (A) -(q), (B) - (s), (C) - (p), (D) - (r)
• C. (A) -(q), (B) - (r), (C) - (s), (D) - (p)
• D. (A) -(r), (B) - (s), (C) - (p), (D) - (q)

Answer 1

Answer: (B)

(A) Let length of common chord be $2a$, then
$\sqrt{9-a^2}+\sqrt{16-a^2}=5$
$\sqrt{16-a^2}=5-\sqrt{9-a^2}$
$16-a^2=25+9-a^2-10\sqrt{9-a^2}$
$10\sqrt{9-a^2}=18\Rightarrow \left(9-a^2\right)=324$
i.e $100a^2=576$
$\therefore a=\sqrt{\frac{576}{100}}=\frac{24}{10}$
$\therefore 2a=\frac{24}{10}=\frac{k}{5}\Rightarrow k=24$
(B) Equation of common chord is $6x+4y+p+q=0$ common chord $(-2,-6)$ pass through centre $x^2+y^2+4x+12y+p=0\therefore p+q=36$
(C) Equation of the circle is $2x^2+2y^2-2\sqrt{2}x-y=0$
Let $(\alpha ,0)$ be mid point of a chord. Then 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 $\left(\sqrt{2},\frac{1}{2}\right)$
$\therefore 2\sqrt{2}\alpha -\sqrt{2}(\sqrt{2}+\alpha )+9=0$
i.e. $(2\sqrt{2}\alpha -3{)}^2=0$
i.e., $\alpha =\frac{3}{2\sqrt{2}},\frac{3}{2\sqrt{2}}$
$\therefore$ Number of chords is 1 .
(D) Mid point of $AB=(1,4)$
$\therefore$ Equation perpendicular bisector of $AB$ is $x=1$
A diameter is $4y=x+7$
$\therefore$ Centre of the circle is $(1,2)$
$\therefore$ Sides of the rectangle are 8 and 4
$\therefore$ area $=32$