Question

State $T$ for true and $F$ for false.
(i) The equation of the circle having centre at ($3,-4)$ and touching the line $5x+12y-12=0$ is ${\left(x-3\right)}^2+{\left(y+4\right)}^2={\left(\frac{45}{13}\right)}^2\cdot$
(ii) The equation of the circle circumscribing the triangle whose sides are the lines $y=x+2$, $3y=4x$, $2y=3x$ is $x^2-y^2-46x+22y=0$.
(iii) The equation of the parabola having focus at $(-1,-2)$ and directrix is $x-2y+3=0$ is $x^2+y^2+3x+32y+4xy+16=0$.

	(i)	(ii)	(iii)
(a)	$F$	$F$	$F$
(b)	$F$	$T$	$F$
(c)	$F$	$F$	$T$
(d)	$T$	$F$	$F$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (D)

$\left(i\right)$ True
The perpendicular distance from centre $\left(3,-4\right)$ to the line $5x+12y-12=0$ is
$d=\left|\frac{15-48-12}{\sqrt{25+144}}\right|=\frac{45}{13}$

$\therefore$ r(radius of the circle) $=\frac{45}{13}$
Hence, the required equation of the circle having centre at $\left(3,-4\right)$ is ${\left(x-3\right)}^2+{\left(y+4\right)}^2={\left(\frac{45}{13}\right)}^2$
$\left(ii\right)$ False
Given equations of lines are
$y=x+\mathrm{2...}\left(i\right),3y=4x...\left(ii\right),2y=3x...\left(iii\right)$
From $\left(i\right)$ and $\left(ii\right)$, $\frac{4x}{3}=x+2$
$\Rightarrow x=6$
Substituting $x=6$ in $\left(ii\right)$, we get $y=8$
$\therefore$ Point $P$ is $\left(6,8\right)$
From $\left(i\right)$ and $\left(iii\right)$, we get
$\frac{3x}{2}=x+2\Rightarrow x=4$

Substituting $x=4$ in $\left(i\right)$, we get $y=6$
$\therefore$ Point $Q$ is $\left(4,6\right)$
From $\left(ii\right)$ and $\left(iii\right)$, $x=0,y=0$
$\therefore$ Point $R$ is $\left(0,0\right)$
Let the equation of the circle is
$x^2+y^2+2gx+2fy+c=0$
Since the points $P\left(6,8\right)$, $Q\left(4,6\right)$ and $R\left(0,0\right)$ lie on the circle.
$\therefore 36+64+12g+16f+c=0$
$\Rightarrow 12g+16f+c=-100\dots \left(iv\right)$
and $16+36+8g+12f+c=0$
$\Rightarrow 8g+12f+c=-52\dots \left(v\right)$
and $c=0\dots \left(vi\right)$
Solving $\left(iv\right)$, $\left(v\right)$ and $\left(vi\right)$, we get
$g=-23,f=11,c=0$
Hence, the required equation of circle is
$x^2+y^2-46x+22y=0$
$\left(iii\right)$ False
We have given, focus $F\left(-1,-2\right)$ and directrix $x-2y+3=0$ If any point on the parabola be $P\left(x,y\right)$, then
$PF=\left|\frac{x-2y+3}{\sqrt{1+4}}\right|$
$\Rightarrow \sqrt{{\left(x+1\right)}^2+{\left(y+2\right)}^2}=\left|\frac{x-2y+3}{\sqrt{5}}\right|$
$\Rightarrow {\left(x+1\right)}^2+{\left(y+2\right)}^2=\frac{{\left(x-2y+3\right)}^2}{5}$
$\Rightarrow 5\left[x^2+2x+1+y^2+4y+4\right]$
$=x^2+4y^2+9-4xy-12y+6x$
$\Rightarrow 4x^2+y^2+4x+32y+4xy+16=0$
which is required equation of parabola.