1. Tardigrade
  2. Question
  3. Mathematics
  4. Column I Column II A The length of the common chord of two circles of radii 3 and 4 units which intersect orthogonally is ( k /5), then k equals to p 1 B The circumference of the circle x2+y2+4 x+12 y+p=0 is bisected by the circlc x2+y2-2 x+8 y-q=0, then p+q is equal to q 24 C Number of distinct chords of the circles 2 x(x-√2)+y(2 y-1)=0 is passing through the point (√2, (1/2)) and are bisected by x-axis is r 32 D One of the diameters of the circles circumscribing the rectangle A B C D is 4 y=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

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Q.
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 $x^2+y^2+4 x+12 y+p=0$ is bisected by the circlc $x^2+y^2-2 x+8 y-q=0$, then $p+q$ is equal to q 24
C Number of distinct chords of the circles $2 x(x-\sqrt{2})+y(2 y-1)=0$ 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 $A B C D$ is $4 y=x+7$. If $A$ and $B$ are the points $(-3,4)$ and $(5,4)$ respectively, then the area of the rectangle is equal to s 36

Conic Sections

Solution:

(A) Let length of common chord be $2 a$, 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 $100 a^2=576 $
$ \therefore a=\sqrt{\frac{576}{100}}=\frac{24}{10} $
$ \therefore 2 a=\frac{24}{10}=\frac{k}{5} \Rightarrow k=24$
(B) Equation of common chord is $6 x+4 y+p+q=0$ common chord $(-2,-6)$ pass through centre $x^2+y^2+4 x+12 y+p=0 \therefore p+q=36$
(C) Equation of the circle is $2 x^2+2 y^2-2 \sqrt{2} x-y=0$
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 $A B=(1,4)$
$\therefore$ Equation perpendicular bisector of $A B$ is $x=1$
A diameter is $4 y = x +7$
$\therefore$ Centre of the circle is $(1,2)$
$\therefore$ Sides of the rectangle are 8 and 4
$\therefore$ area $=32$