Let lengths of intercepts on x -axis and y -axis made by circle x^{2}+y^{2}+a x+2 a y+c=0 (a

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10-1.Circle and System of Circles

hard

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

$\sqrt{11}$

$\sqrt{7}$

$\sqrt{6}$

$\sqrt{10}$

(JEE MAIN-2021)

Solution

$x ^{2}+ y ^{2}+ ax +2 ay + c =0$

$2 \sqrt{ g ^{2}- c }=2 \sqrt{\frac{ a ^{2}}{4}- c }=2 \sqrt{2}$

$\Rightarrow \quad \frac{ a ^{2}}{4}- c =2…….(1)$

$2 \sqrt{ f ^{2}- c }=2 \sqrt{ a ^{2}- c }=2 \sqrt{5}$

$\Rightarrow a^{2}-c=5……(2)$

$(1)$ and $(2)$

$\frac{3 a ^{2}}{4}=3 \Rightarrow a =-2 \quad( a < 0)$

$\therefore \quad c=-1$

Circle $\Rightarrow x^{2}+y^{2}-2 x-4 y-1=0$

$\Rightarrow(x-1)^{2}+(y-2)^{2}=6$

Given $x+2 y=0 \Rightarrow m=-\frac{1}{2}$

$m _{\text {tangent }}=2$

Equation of tangent

$\Rightarrow(y-2)=2(x-1) \pm \sqrt{6} \sqrt{1+4}$

$\Rightarrow 2 x-y \pm \sqrt{30}=0$

Perpendicular distance from $(0,0)=\left|\frac{\pm \sqrt{30}}{\sqrt{4+1}}\right|=\sqrt{6}$

Standard 11

Mathematics

Two concentric circles are such that the smaller divides the larger into two regions of equal area. If the radius of the smaller circle is $2$ , then the length of the tangent from any point $' P '$ on the larger circle to the smaller circle is :

normal

The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is

easy

The point of contact of the tangent to the circle ${x^2} + {y^2} = 5$ at the point $(1, -2)$ which touches the circle ${x^2} + {y^2} – 8x + 6y + 20 = 0$, is

hard

The equation of the circle whose radius is $5$ and which touches the circle ${x^2} + {y^2} – 2x – 4y – 20 = 0$ externally at the point $(5, 5)$ is

medium

(IIT-1979)

If the line $x = k$ touches the circle ${x^2} + {y^2} = 9$, then the value of $k$ is