Question

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

Answer 1

Hint: 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 $(-g,-f)$ . 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 $(-g,-f)$ 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,

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

$\Rightarrow g=-{\displaystyle \frac{1}{2}}$

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 $g=-{\displaystyle \frac{1}{2}}\text{ and }c=0.$

So, we have $C_1{C}_2=3\pm \sqrt{{(-{\displaystyle \frac{1}{2}})}^2+f^2-0}$

$\Rightarrow C_1{C}_2=3\pm \sqrt{{\displaystyle \frac{1}{4}}+f^2}...\left(i\right)$

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 $({\displaystyle \frac{1}{2}},-f)$ . So, by distance formula, we can write

$C_1{C}_2=\sqrt{{(0-{\displaystyle \frac{1}{2}})}^2+{(0+f)}^2}$

$\Rightarrow C_1{C}_2=\sqrt{{\displaystyle \frac{1}{4}}+f^2}...\left(ii\right)$

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

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

Or, we can write

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

If we take the negative sign, we will have

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

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

Taking the positive sign, we get

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

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

Squaring both sides, we get

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

We can also write the above as

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

$\Rightarrow f^2=2$

Taking square root on both sides, we get

$f=\pm \sqrt{2}$

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

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.