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 $x^{2} - y^{2} - 46x + 2 2y = 0$.
(iii) The equation of the parabola having focus at $(-1, -2)$ and directrix is $x - 2 y + 3 = 0$ is $x^{2 }+y^{2} + 3x +32y + 4 xy + 1 6 = 0$.

	(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
$y = x + 2 ...\left(i\right), 3y = 4x ...\left(ii\right), 2y = 3x ...\left(iii\right)$
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.
$\therefore 36 + 64+ 12g+16f+c = 0$
$\Rightarrow 12g+ 16f+ c = -100\, \ldots\left(iv\right)$
and $16 + 36 + 8g + 12f+ c = 0$
$\Rightarrow 8g + 12f+ c = -52\, \ldots\left(v\right)$
and $c = 0 \,\ldots\left(vi\right)$
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.