Let the lengths of intercepts on the x-axis a | vedclass

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

Question

(C) The equation of the circle is $x^{2}+y^{2}+ax+2ay+c=0$.The length of the $x$-intercept is $2\sqrt{g^{2}-c} = 2\sqrt{\frac{a^{2}}{4}-c} = 2\sqrt{2}

Vedclass · Accepted Answer

$\sqrt{6}$

Vedclass · Answer

(C) The equation of the circle is $x^{2}+y^{2}+ax+2ay+c=0$.The length of the $x$-intercept is $2\sqrt{g^{2}-c} = 2\sqrt{\frac{a^{2}}{4}-c} = 2\sqrt{2}$.$\Rightarrow \frac{a^{2}}{4}-c = 2 \quad \dots(1)$The length of the $y$-intercept is $2\sqrt{f^{2}-c} = 2\sqrt{a^{2}-c} = 2\sqrt{5}$.$\Rightarrow a^{2}-c = 5 \quad \dots(2)$Subtracting $(1)$ from $(2)$:$(a^{2}-c) - (\frac{a^{2}}{4}-c) = 5-2$ $\Rightarrow \frac{3a^{2}}{4} = 3$ $\Rightarrow a^{2} = 4$.Since $a Substituting $a = -2$ into $(2)$: $(-2)^{2}-c = 5$ $\Rightarrow 4-c = 5$ $\Rightarrow c = -1$.The circle equation is $x^{2}+y^{2}-2x-4y-1 = 0$,which is $(x-1)^{2}+(y-2)^{2} = 6$.The center is $(1, 2)$ and the radius $r = \sqrt{6}$.The tangent is perpendicular to $x+2y=0$,so its slope $m = 2$.The equation of the tangent is $(y-2) = 2(x-1) \pm \sqrt{6}\sqrt{1+2^{2}}$.$y-2 = 2x-2 \pm \sqrt{30} \Rightarrow 2x-y \pm \sqrt{30} = 0$.The distance from the origin $(0,0)$ to the tangent $2x-y \pm \sqrt{30} = 0$ is $d = \frac{|\pm \sqrt{30}|}{\sqrt{2^{2}+(-1)^{2}}} = \frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$.

Let the lengths of intercepts on the $x$-axis and $y$-axis made by the circle $x^{2}+y^{2}+ax+2ay+c=0$ $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$,respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$ is equal to:

Similar Questions

If the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = 16$,then $m =$

If the straight line $y = mx + c$ touches the circle ${x^2} + {y^2} - 4y = 0$,then the value of $c$ will be

The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals and passing through the point $(2, 1)$ is:

The line $y = x + a\sqrt{2}$ is a tangent to the circle $x^2 + y^2 = a^2$ at which of the following points?

If the straight line $4x + 3y + \lambda = 0$ touches the circle $2(x^2 + y^2) = 5$,then $\lambda$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module