Question

The center of the circle passing through (0, 0) and (1, 0) and touching the circle ${{x}^{2}}+{{y}^{2}}=9$ is:
(a) $\left( \dfrac{1}{2},\dfrac{1}{2} \right)$
(b) $\left( \dfrac{1}{2},-\sqrt{2} \right)$
(c) $\left( \dfrac{3}{2},\dfrac{1}{2} \right)$
(d) $\left( \dfrac{1}{2},\dfrac{3}{2} \right)$

Answer 1

Hint: To solve this question, we will consider the general equation of the circle and with the help of the given conditions, we will find the value of g, f and c. And then we will be able to choose 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 ${{x}^{2}}+{{y}^{2}}=9$. 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,
$0+0+2g\left( 0 \right)+2f\left( 0 \right)+c=0$
$c=0....\left( i \right)$
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 ${{x}^{2}}+{{y}^{2}}=9$. And we know that in such cases, the distance between the centers is equal to the sum or the difference of the radius. For that, we will find the coordinates of the center of both the circle and the radius of both the circle and then put the values in
${{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}$
We know that the center of any circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is (– g, – f). So, for the circle ${{x}^{2}}+{{y}^{2}}=9$, the center is C’ (0, 0) and for ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, the center is C (– g, – f) and we know that $g=\dfrac{-1}{2}$. Therefore, the coordinates of C is $\left( \dfrac{1}{2},-f \right)$.
We also know that the radius of the circle is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. So, the radius
$r=\sqrt{{{\left( -\dfrac{1}{2} \right)}^{2}}+{{f}^{2}}-0}$
$r=\sqrt{\dfrac{1}{4}+{{f}^{2}}}$
We know that r’ = 3. Using the distance formula, we can say that,
$CC'=\sqrt{{{\left( 0-\dfrac{1}{2} \right)}^{2}}+{{\left( 0+f \right)}^{2}}}$
$CC'=\sqrt{\dfrac{1}{4}+{{f}^{2}}}$
So, we get the equation $CC'={{r}_{1}}\pm {{r}_{2}}$ as
$\sqrt{\dfrac{1}{4}+{{f}^{2}}}=3\pm \sqrt{\dfrac{1}{4}+{{f}^{2}}}$
$\sqrt{\dfrac{1}{4}+{{f}^{2}}}\mp \sqrt{\dfrac{1}{4}+{{f}^{2}}}=3$
Here, we will ignore negative sign because that will cancel $\sqrt{\dfrac{1}{4}+{{f}^{2}}}$. So, we can write,
$\sqrt{\dfrac{1}{4}+{{f}^{2}}}+\sqrt{\dfrac{1}{4}+{{f}^{2}}}=3$
$2\sqrt{\dfrac{1}{4}+{{f}^{2}}}=3$
$\sqrt{\dfrac{1}{4}+{{f}^{2}}}=\dfrac{3}{2}$
Now, we will square both the sides of the equations,
$\dfrac{1}{4}+{{f}^{2}}=\dfrac{9}{4}$
${{f}^{2}}=\dfrac{9}{4}-\dfrac{1}{4}$
${{f}^{2}}=\dfrac{8}{4}$
${{f}^{2}}=2$
$f=\pm \sqrt{2}$
Hence, we get the value as $f=\pm \sqrt{2}$. So, we can say that the center of ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is $\left( \dfrac{1}{2},\pm \sqrt{2} \right)$.
Therefore, option (b) is the right answer.

Note: The possible mistake one can make in this question is by putting ${{x}^{2}}+{{y}^{2}}=9$ 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.Students should remember general equation of circle and centre of any circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is (– g, – f) and the radius of the circle is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.