Question

The circle ${{x}^{2}}+{{y}^{2}}+2{{a}_{1}}x+c=0$ lies completely inside the circle ${{x}^{2}}+{{y}^{2}}+2{{a}_{2}}x+c=0$, then:
(a) ${{a}_{1}}{{a}_{2}}>0,c<0$
(b) ${{a}_{1}}{{a}_{2}}>0,c>0$
(c) ${{a}_{1}}{{a}_{2}}<0,c<0$
(d) ${{a}_{1}}{{a}_{2}}<0,c>0$

Answer 1

Hint: To solve this question, we should know that whenever a point lies in the circle of some equation, then for that point the equation of the circle becomes less than 0. Also, we should know that for the equation of the circle, ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, the center is (– g, – f) and the radius is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.

Complete step-by-step answer:
To solve this question, we have been given that the circle ${{x}^{2}}+{{y}^{2}}+2{{a}_{1}}x+c=0$ lies completely inside the circle ${{x}^{2}}+{{y}^{2}}+2{{a}_{2}}x+c=0$. Let us consider ${{C}_{1}}$ as ${{x}^{2}}+{{y}^{2}}+2{{a}_{1}}x+c=0$ and ${{C}_{2}}$ as ${{x}^{2}}+{{y}^{2}}+2{{a}_{2}}x+c=0$. And we know that for the equation of the circle, ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, the center is (– g, – f). So, we can say that the center of the circle ${{C}_{1}}$ is at $\left( -{{a}_{1}},0 \right)$ and that of the circle ${{C}_{2}}$ is at $\left( -{{a}_{2}},0 \right)$. And we have been given that ${{C}_{1}}$ completely lies inside ${{C}_{2}}$, then we can say for $\left( -{{a}_{1}},0 \right)$, ${{x}^{2}}+{{y}^{2}}+2{{a}_{2}}x+c<0$. So, after putting the values, we will get,
${{\left( -{{a}_{1}} \right)}^{2}}+{{\left( 0 \right)}^{2}}+2{{a}_{2}}\left( -{{a}_{1}} \right)+c<0$
${{\left( {{a}_{1}} \right)}^{2}}-2{{a}_{2}}{{a}_{1}}+c<0.....\left( i \right)$
We also know that the radius of the circle of the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. So, the radius of ${{C}_{1}}$ is $\sqrt{{{\left( {{a}_{1}} \right)}^{2}}-c}$ and radius of ${{C}_{2}}$ is $\sqrt{{{\left( {{a}_{2}} \right)}^{2}}-c}$. And we know that ${{C}_{1}}$ lies inside ${{C}_{2}}$. So, we can write
The radius of ${{C}_{1}}$ < Radius of ${{C}_{2}}$
$\Rightarrow \sqrt{{{\left( {{a}_{1}} \right)}^{2}}-c}<\sqrt{{{\left( {{a}_{2}} \right)}^{2}}-c}$
Now, we will square both sides of the equation. So, we get,
$a_{1}^{2}-{c}<{a_{2}^{2}}-c$
${a_{1}^{2}}<{a}_{2}^{2}$
${{a}_{1}}<{{a}_{2}}.....\left( ii \right)$
Now, we will write equation (i) as
$c<2{{a}_{1}}{{a}_{2}}-a_{1}^{2}$
As ${{a}_{1}}<{{a}_{2}}$, therefore we can say ${{\left( {{a}_{1}} \right)}^{2}}<{{a}_{1}}{{a}_{2}}$ which means ${{a}_{1}}{{a}_{2}}-a_{1}^{2}>0$ and from this we can conclude that c > 0. We can also write the equation (i) as,
$a_{1}^{2}+c<2{{a}_{1}}{{a}_{2}}$
As we proved c > 0, then $a_{1}^{2}+c$ will definitely be greater than 0. So, we can write
${0}<{a_{1}^{2}}+{c}<{2{{a}_{1}}{{a}_{2}}}$
${{a}_{1}}{{a}_{2}}>0$
Hence, we can say ${{a}_{1}}{{a}_{2}}>0$ and c > 0.
Therefore, option (b) is the right answer.

Note: The possible mistake one can make is after the equation (i), that is by not taking relation between radius and then making silly assumptions that might give you the correct answer but that will be a random answer. To get a confirmed answer, we need to know the relation between their radius.