Question

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

Answer 1

We need to find the centre of the circle. So, we can assume the general equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ to be its equation with centre $\left( -g,-f \right)$ . Now, we should solve for g, f and c to solve the problem. We must remember that when two circles touch each other then ${{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}$.

Complete step-by-step solution:

Let the equation of the required circle be ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ . We know that the centre of this circle is $\left( -g,-f \right)$ and the radius of this circle is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ .

We are given that this circle passes through the point (0,0). So, substituting this point in the equation, we get

$c=0$

Thus, we have the equation as ${{x}^{2}}+{{y}^{2}}+2gx+2fy=0$ .

The point (1,0) also lies on this circle. So, substituting this point in the equation we get,

$1+0+2g+0=0$

$\Rightarrow g=-\dfrac{1}{2}$

Substituting this value in our equation we get

${{x}^{2}}+{{y}^{2}}-x+2fy=0$

We are given that this circle touches the circle ${{x}^{2}}+{{y}^{2}}=9$ . We know that in such cases the distance between the centres is equal to the algebraic sum or difference of their radii.

Thus, we have ${{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}$ .

We can clearly see that the radius of ${{x}^{2}}+{{y}^{2}}=9$ is 3. And the radius of ${{x}^{2}}+{{y}^{2}}-x+2fy=0$ is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ where $g=-\dfrac{1}{2}\text{ and }c=0.$

So, we have ${{C}_{1}}{{C}_{2}}=3\pm \sqrt{{{\left( -\dfrac{1}{2} \right)}^{2}}+{{f}^{2}}-0}$

$\Rightarrow {{C}_{1}}{{C}_{2}}=3\pm \sqrt{\dfrac{1}{4}+{{f}^{2}}}...\left( i \right)$

We know that the centre of ${{x}^{2}}+{{y}^{2}}=9$ is (0,0) and the centre of ${{x}^{2}}+{{y}^{2}}-x+2fy=0$ is $\left( \dfrac{1}{2},-f \right)$ . So, by distance formula, we can write

${{C}_{1}}{{C}_{2}}=\sqrt{{{\left( 0-\dfrac{1}{2} \right)}^{2}}+{{\left( 0+f \right)}^{2}}}$

$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{\dfrac{1}{4}+{{f}^{2}}}...\left( ii \right)$

Equating equation (i) and (ii), we get

$\sqrt{\dfrac{1}{4}+{{f}^{2}}}=3\pm \sqrt{\dfrac{1}{4}+{{f}^{2}}}$

Or, we can write

$\sqrt{\dfrac{1}{4}+{{f}^{2}}}\mp \sqrt{\dfrac{1}{4}+{{f}^{2}}}=3$

If we take the negative sign, we will have

$\sqrt{\dfrac{1}{4}+{{f}^{2}}}-\sqrt{\dfrac{1}{4}+{{f}^{2}}}=3$

$\Rightarrow 0=3$ which makes no sense. So, we can neglect the negative sign.

Taking the positive sign, we get

$\sqrt{\dfrac{1}{4}+{{f}^{2}}}+\sqrt{\dfrac{1}{4}+{{f}^{2}}}=3$

$\Rightarrow \sqrt{\dfrac{1}{4}+{{f}^{2}}}=\dfrac{3}{2}$

Squaring both sides, we get

$\dfrac{1}{4}+{{f}^{2}}=\dfrac{9}{4}$

We can also write the above as

${{f}^{2}}=\dfrac{9}{4}-\dfrac{1}{4}$

$\Rightarrow {{f}^{2}}=2$

Taking square root on both sides, we get

$f=\pm \sqrt{2}$

So, the centre of this circle is either $\left( \dfrac{1}{2},-\sqrt{2} \right)\text{ or }\left( \dfrac{1}{2},\sqrt{2} \right).$

Hence, option (d) is the correct answer.

Note: Some students may try to solve this problem by calculating c and f in the same manner, and for g, solving ${{x}^{2}}+{{y}^{2}}=9$ and ${{x}^{2}}+{{y}^{2}}-x+2fy=0$ simultaneously. In this method, we will still have x and y is the equation and won’t be able to eliminate them. So, we should use the same concept as above to find f.