The straight line touching the circle x^2+y^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The line is normal to the circle $x^2+y^2-4y-6=0$,so it must pass through its center $(0, 2)$.Let the slope of the line be $m$. The equation of th

Vedclass · Accepted Answer

$4x-3y+6=0$

Vedclass · Answer

(A) The line is normal to the circle $x^2+y^2-4y-6=0$,so it must pass through its center $(0, 2)$.Let the slope of the line be $m$. The equation of the line is $y-2=m(x-0)$,which simplifies to $mx-y+2=0$.This line is tangent to the circle $x^2+y^2-2x-3=0$,which has center $(1, 0)$ and radius $r = \sqrt{1^2+0^2-(-3)} = 2$.The perpendicular distance from the center $(1, 0)$ to the line $mx-y+2=0$ must equal the radius $2$.$\frac{|m(1)-0+2|}{\sqrt{m^2+(-1)^2}} = 2$$|m+2| = 2\sqrt{m^2+1}$Squaring both sides: $(m+2)^2 = 4(m^2+1)$$m^2+4m+4 = 4m^2+4$$3m^2-4m = 0$$m(3m-4) = 0$So,$m=0$ or $m=\frac{4}{3}$.If $m=0$,the line is $y-2=0(x)$,which is $y=2$.If $m=\frac{4}{3}$,the line is $y-2=\frac{4}{3}x$,which is $3y-6=4x$,or $4x-3y+6=0$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

The straight line touching the circle $x^2+y^2-2x-3=0$ and remaining normal to the circle $x^2+y^2-4y-6=0$ is

Similar Questions

If the point of intersection of the pair of the transverse common tangents and that of the pair of direct common tangents drawn to the circles $x^2+y^2-14x+6y+33=0$ and $x^2+y^2+30x-2y+1=0$ are $T$ and $D$ respectively,then the centre of the circle having $TD$ as diameter is

Let $AB$ be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then the locus of the centroid of the $\Delta PAB$ as $P$ moves on the circle is

$A$ circle touches the parabola $y^2=4x$ at $(1,2)$ and also touches its directrix. The $y$-coordinate of the point of contact of the circle and the directrix is

Vedclass Test Series

Exam Paper Generator

Online Exam Module