Question

The center of the circle(s) passing through the points (0, 0), (1, 0) and touching the circle ${{x}^{2}}+{{y}^{2}}=16$, is(are)
(a) $\left( \dfrac{1}{2},\dfrac{\sqrt{15}}{2} \right)$
(b) $\left( \dfrac{1}{2},-\dfrac{\sqrt{15}}{2} \right)$
(c) $\left( \dfrac{1}{2},{{2}^{\dfrac{1}{2}}} \right)$
(d) $\left( \dfrac{1}{2},-{{2}^{\dfrac{1}{2}}} \right)$

Answer 1

To solve this question, we will consider the general equation of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ and with the help of given conditions, we will find the value of g, f, and c. And then we will put the value of g, f, and c to get the correct answer.

Complete step by step answer:
In this question, we have to find the center of the circle which passes through (0, 0) and (1, 0) and touches the circle ${{x}^{2}}+{{y}^{2}}=16$. Now, let us consider the equation of the circle of which we have to find the center, ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. Now, by using the given conditions, we will find the value of g, f, and c to find the center of the circle. We have been given that the circle passes through (0, 0). So, we will put the value of x = 0 and y = 0 to get the value of c. So, we get,
${{\left( 0 \right)}^{2}}+{{\left( 0 \right)}^{2}}+2g\left( 0 \right)+2f\left( 0 \right)+c=0$
$c=0....\left( i \right)$
Now, we can write the equation of the circle as
${{x}^{2}}+{{y}^{2}}+2gx+2fy=0$
Now, we have been also given that the circle passes through (1, 0). So, we will put x = 1 and y = 0 in the above equation of the circle. Therefore, we get,
${{\left( 1 \right)}^{2}}+0+2g\left( 1 \right)+2f\left( 0 \right)=0$
$1+2g=0$
$g=-\dfrac{1}{2}....\left( ii \right)$
Now, we have been given that ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ touches ${{x}^{2}}+{{y}^{2}}=16$. And we know that in such cases the distance between the center is equal to the algebraic sum of the radius. Mathematically, we can represent it as,
${{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}....\left( iii \right)$
We know that the center of any circle of the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is given by (– g, – f) and radius is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. So, the center and radius of ${{x}^{2}}+{{y}^{2}}+2gx+2fy=0$ are C = (– g, – f) and $r=\sqrt{{{g}^{2}}+{{f}^{2}}}$ respectively. And the center and radius of ${{x}^{2}}+{{y}^{2}}=16$ are C’ (0, 0) and r’ = 4 respectively.
Now, we know that $g=\dfrac{-1}{2}$. Therefore, we can say $C\left( \dfrac{1}{2},-f \right)$ , and $r=\sqrt{\dfrac{1}{4}+{{f}^{2}}}$. So, we can say that $CC'=\sqrt{{{\left( 0-\dfrac{1}{2} \right)}^{2}}+{{\left( 0+f \right)}^{2}}}$ by distance formula, that is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$CC'=\sqrt{\dfrac{1}{4}+{{f}^{2}}}$
Now, we will put the values in equation (iii), So, we will get,
$\sqrt{\dfrac{1}{4}+{{f}^{2}}}=4\pm \sqrt{\dfrac{1}{4}+{{f}^{2}}}$
$\sqrt{\dfrac{1}{4}+{{f}^{2}}}\mp \sqrt{\dfrac{1}{4}+{{f}^{2}}}=4$
Here, we will ignore the negative sign. So, we will get,
$2\sqrt{\dfrac{1}{4}+{{f}^{2}}}=4$
Now, we will square both sides of the equation. So, we will get,
$4\left[ \dfrac{1}{4}+{{f}^{2}} \right]=16$
$\dfrac{1}{4}+{{f}^{2}}=\dfrac{16}{4}$
${{f}^{2}}=4-\dfrac{1}{4}$
${{f}^{2}}=\dfrac{15}{4}$
$f=\pm\dfrac{\sqrt{15}}{2}$
So, we can write the center(s) of the circle which passes through (0, 0) and (1, 0) and touches ${{x}^{2}}+{{y}^{2}}=16$ as $\left( \dfrac{1}{2},\pm\dfrac{\sqrt{15}}{2} \right)$.

Therefore, option (a) and (b) are the correct answers.

Note: The possible mistake one can make in this question is by putting ${{x}^{2}}+{{y}^{2}}=16$ in ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. Because in that, we will be having x and y in the equation and then we won’t be able to find the value of c. Also, one can get confused between option (a) and (b), that which is correct but actually, they both are correct