The equation of a common tangent to the circl | vedclass

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

Question

(B) Let the equation of the common tangent be $y=mx+c$. For the circle $x^2+y^2=4$,the condition for tangency is $c^2 = r^2(1+m^2)$,where $r^2=4$. So,

Vedclass · Accepted Answer

$\sqrt{2}x-\sqrt{21}y+2\sqrt{23}=0$

Vedclass · Answer

(B) Let the equation of the common tangent be $y=mx+c$. For the circle $x^2+y^2=4$,the condition for tangency is $c^2 = r^2(1+m^2)$,where $r^2=4$. So,$c^2 = 4(1+m^2)$. For the ellipse $\frac{x^2}{25} + \frac{y^2}{2} = 1$,the condition for tangency is $c^2 = a^2m^2 + b^2$,where $a^2=25$ and $b^2=2$. So,$c^2 = 25m^2 + 2$. Equating the two expressions for $c^2$: $4(1+m^2) = 25m^2 + 2$ $4 + 4m^2 = 25m^2 + 2$ $21m^2 = 2$ $\Rightarrow m^2 = \frac{2}{21}$ $\Rightarrow m = \pm \sqrt{\frac{2}{21}}$. Substituting $m^2$ back into $c^2 = 4(1+m^2)$: $c^2 = 4(1 + \frac{2}{21}) = 4(\frac{23}{21}) = \frac{92}{21}$. Thus,$c = \pm \sqrt{\frac{92}{21}} = \pm \frac{2\sqrt{23}}{\sqrt{21}}$. The equation of the tangent is $y = \pm \sqrt{\frac{2}{21}}x \pm \frac{2\sqrt{23}}{\sqrt{21}}$. Multiplying by $\sqrt{21}$: $\sqrt{21}y = \pm \sqrt{2}x \pm 2\sqrt{23}$. Rearranging gives $\sqrt{2}x - \sqrt{21}y + 2\sqrt{23} = 0$.

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 equation of a common tangent to the circle $x^2+y^2=4$ and the ellipse $2x^2+25y^2=50$ is

Similar Questions

Let $y=mx+c, m>0$ be the focal chord of $y^{2}=-64x$,which is tangent to $(x+10)^{2}+y^{2}=4$. Then,the value of $4\sqrt{2}(m+c)$ is equal to $.....$

The angle at which the circles $(x - 1)^2 + y^2 = 10$ and $x^2 + (y - 2)^2 = 5$ intersect is

The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module